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Class 9 Mathematics Chapter 10 of 15

Chapter 10 — Circles

Overview

This chapter introduces the circle as a fundamental plane figure and develops the basic language, properties and simple theorems about chords, arcs and angles. It explains centre, radius, diameter, chord, arc, sector and segment, and then builds key geometric results: how chords relate to the centre (equal chords, distances from the centre, perpendicular bisectors) and how an arc/chord subtends angles at the centre and at points on the circle. The chapter trains students to read and write precise geometric statements, construct simple figures with ruler and compass, prove short theorems, and apply these results to solve problems. Understanding these ideas is important because circles and their properties recur across geometry, problem solving and later topics (constructions, tangents, cyclic quadrilaterals).

Learning Objectives

  • Define circle and related terms: radius, diameter, chord, arc, sector, segment, tangent and secant.
  • State and prove basic theorems on equal chords and equal arcs, and the fact that the perpendicular from the centre to a chord bisects the chord and its arc.
  • Prove that the angle subtended by a chord at the centre is twice the angle subtended at the circumference and that an angle in a semicircle is a right angle.
  • Apply the property that the radius to a tangent is perpendicular to the tangent and that tangents from an external point are equal to solve problems.
  • Construct a tangent to a circle at a given point and construct the perpendicular bisector of a chord to locate the centre using ruler and compass.
  • Solve numerical problems to determine lengths of chords, radii and distances in circle configurations using circle theorems and Pythagoras.
  • Calculate lengths of arcs and areas of sectors and segments using degree-based formulae and appropriate unit conversion.
  • Prove and apply properties of angles formed by two chords, a chord and a tangent, and two tangents to find unknown angles.

Topics in this chapter

10 topics · tap a topic title to jump straight to it.

🔢1

Basic definitions and terminology

A circle is the set of all points in a plane that are at a fixed distance from a fixed point. The fixed point is called the centre and the fixed distance is called the radius.

Key terms and their meanings:

  • Centre (O): The fixed point from which all points on the circle are equidistant.
  • Radius (r): The distance from the centre to any point on the circle. Every radius has the same length.
  • Diameter (d): A line segment passing through the centre whose endpoints lie on the circle. d = 2r.
  • Circumference: The length or perimeter of the circle (complete boundary).
  • Chord: A line segment with both endpoints on the circle. A diameter is a chord that passes through the centre.
  • Arc: A continuous part of the circle's circumference between two points. A minor arc is the shorter arc between the points, a major arc is the longer one.
  • Semicircle: An arc equal to half the circumference (endpoints of a diameter).
  • Quadrant: One fourth of a circle (90° sector).
  • Sector: The region bounded by two radii and the included arc (like a pizza slice). Its size is measured by the central angle.
  • Segment: The region bounded by a chord and the corresponding arc (area of sector minus area of triangle formed by the radii).
  • Tangent: A line that touches the circle at exactly one point. The radius drawn to the point of contact is perpendicular to the tangent.
  • Secant: A line that intersects the circle at two points.
  • Concentric circles: Two or more circles that share the same centre but have different radii.
  • Central angle: An angle whose vertex is at the centre and whose sides are radii; it subtends an arc.

Simple properties to remember (useful for later proofs):

  • The perpendicular from the centre to a chord bisects the chord and its arcs.
  • Equal chords of a circle subtend equal arcs and are equidistant from the centre.
  • A tangent is perpendicular to the radius at the point of contact.
📌 Examples
  • Wheel of a bicycle: the rim is a circle; the hub is the centre; spokes are radii.
  • A coin: its edge is the circumference; the diameter is the longest straight distance across the coin.
  • A pizza: cutting from the centre produces sectors; a single slice is a sector.
  • A roundabout: lanes form concentric circles if there are multiple circular paths.
  • Ripples on water: concentric circles expanding from the point where a drop fell.
🧮 Formulas
  1. Relationship between diameter and radius: d = 2r
  2. Circumference (perimeter) of a circle: C = 2πr = πd
  3. Area of a circle: A = πr²
  4. Length of an arc (for central angle θ in degrees): arc length = (θ/360) × 2πr
  5. Area of a sector (central angle θ in degrees): sector area = (θ/360) × πr²
  6. Area of a segment = area of sector − area of triangle (formed by the two radii and chord)
📊 Visual ideas
Draw a circle with centre O. Mark a point A on the circle and draw OA to show the radius r. Draw a line through O that meets the circle at B and C to show diameter BC (label d = 2r). Use different colors for radius and diameter.
Sketch a chord AB that does not pass through O. Draw the perpendicular from O to chord AB and show that it bisects AB (label midpoint M). Shade the two arcs determined by A and B and label them 'minor arc AB' and 'major arc A—B'.
Draw two radii OA and OB making a central angle ∠AOB = θ. Shade the sector AOB and label the sector area formula (θ/360) × πr². On the same diagram, mark the arc AB and label its length (θ/360) × 2πr.
Draw a tangent line at point T on the circle. Show radius OT and draw a right-angle symbol at T to indicate OT ⟂ tangent. Label the tangent point as 'point of contact'.
⭕2

Concentric circles and related concepts

Definition: Concentric circles are two or more circles that have the same centre but different radii. They lie one inside another and do not intersect.

Key properties:

  • All concentric circles share a common centre O. If radii are R and r (R > r), the circles are represented by (x - h)^2 + (y - k)^2 = R^2 and (x - h)^2 + (y - k)^2 = r^2.
  • Concentric circles never meet (no common point) when radii are different.
  • There is no common tangent line that touches two distinct concentric circles at the same point; in fact no single straight line can be tangent to both circles simultaneously.
  • If two concentric circles subtend the same central angle θ, lengths of their corresponding arcs are proportional to their radii (arc length = rθ), and areas of corresponding sectors are proportional to the square of the radii (sector area = 1/2 r^2 θ).
  • The region between two concentric circles is called an annulus (a ring). Its area equals the difference of the areas of the larger and smaller circles: π(R^2 − r^2).

Geometric consequences and uses: Because all radii from the centre to the circles are collinear for a given direction, the shortest distance between the two circles along a ray from the centre is |R − r|. Concentric circles appear often in problems that compare lengths or areas of arcs, sectors and rings.

Algebraic form / family of circles: For centre (h,k), the family of concentric circles is (x − h)^2 + (y − k)^2 = r^2, with r varying. In the plane, concentric circles can be used to model levels of equal distance from a point (level curves).

📌 Examples
  • Example 1 (Equations): Show that x^2 + y^2 = 25 and x^2 + y^2 = 9 are concentric. Explanation: both have centre (0,0) with radii 5 and 3 respectively; they do not meet and form an annulus.
  • Example 2 (Area of annulus): Find area between circles with equations (x − 2)^2 + (y + 1)^2 = 36 and (x − 2)^2 + (y + 1)^2 = 4. Solution: R = 6, r = 2 so area = π(36 − 4) = 32π square units.
  • Example 3 (Arc lengths & sectors): Two concentric circles with radii 8 cm and 5 cm subtend an angle of 60° (π/3 rad) at the centre. Corresponding arc lengths are 8·(π/3)=8π/3 cm and 5·(π/3)=5π/3 cm. Sector areas are (1/2)·8^2·(π/3)= (32π/3) cm^2 and (1/2)·5^2·(π/3)=(25π/6) cm^2; the area of the ring-sector between them = (32π/3) − (25π/6) = (39π/6) = 13π/2 cm^2.
🧮 Formulas
  1. Equation of circle with centre (h,k): (x − h)^2 + (y − k)^2 = r^2
  2. Concentric circles: (x − h)^2 + (y − k)^2 = R^2 and (x − h)^2 + (y − k)^2 = r^2 (same h,k; R ≠ r)
  3. Circumference of a circle: C = 2πr
  4. Area of a circle: A = πr^2
  5. Area of an annulus (region between two concentric circles): A = π(R^2 − r^2) = π(R − r)(R + r)
  6. Arc length subtending angle θ (in radians): s = rθ
📊 Visual ideas
Plot two concentric circles with centre (0,0): x^2 + y^2 = 25 (radius 5) and x^2 + y^2 = 9 (radius 3). Shade the annulus between them to show the ring-shaped region.
Plot a family of concentric circles x^2 + y^2 = 1, 4, 9, 16 to illustrate multiple rings and increasing radii.
Plot concentric circles with centre (2,−1): (x − 2)^2 + (y + 1)^2 = 4 and (x − 2)^2 + (y + 1)^2 = 16, and draw a ray from the centre making angle 45°. Mark the two intersection points of the ray with the circles to show |R − r| as the radial gap.
Plot sectors on two concentric circles: draw radii at angles 0 and 60° and shade corresponding sectors on both circles to demonstrate proportionality of arc lengths and the ring-sector area. (Use θ = π/3 radians.)
📐3

Central and inscribed angles

Definition: A central angle of a circle is an angle whose vertex is the centre of the circle and whose sides (arms) are radii. If O is the centre and A, B are points on the circle, then ∠AOB is a central angle.

A inscribed angle (or angle subtended at the circumference) is an angle whose vertex lies on the circle and whose sides pass through two other points on the circle. If C is on the circle and A, B are two other points on the circle, then ∠ACB is an inscribed angle subtending arc AB.

Key relationship (central–inscribed angle theorem): An inscribed angle is half the central angle subtending the same arc. In symbols, if ∠AOB is the central angle and ∠ACB is an inscribed angle subtending the same arc AB, then ∠ACB = (1/2) ∠AOB.

Reason (sketch of proof): Join OA, OB and OC (radii). Triangles OCA and OCB are isosceles (OA = OC = OB). Using base angles in these isosceles triangles and angle-chasing, one gets ∠AOB = 2·∠ACB. Thus the inscribed angle equals half the central angle subtending the same arc.

Other important results:

  • Angles subtending the same arc (or chord) at the circumference are equal. That is, if points C and D lie on the same arc AB, then ∠ACB = ∠ADB.
  • Angle in a semicircle is a right angle (Thales' theorem): an angle subtended by a diameter is 90°.
  • Opposite angles of a cyclic quadrilateral are supplementary: for a quadrilateral ABCD inscribed in a circle, ∠A + ∠C = 180° and ∠B + ∠D = 180°.

These properties are widely used in solving geometry problems involving circles, construction of angles, and proving equalities or perpendicularities.

📌 Examples
  • Clock face: The angle between two hands at the centre is a central angle; if you stand on the rim of a circular track and look at two fixed points on the track, the angle you see between them is an inscribed angle.
  • Pizza slices: Cutting along radii produces central angles — the size of the slice is given by the central angle at the centre of the pizza.
  • Viewing a stadium goal from different seats on the same arc: all spectators on that arc see the goal under the same inscribed angle.
  • A right triangle inscribed with its hypotenuse as diameter (e.g., endpoints of the diameter are two vertices): the angle opposite the diameter is 90° (angle in a semicircle).
🧮 Formulas
  1. If ∠AOB is the central angle and ∠ACB is the inscribed angle subtending the same arc AB, then ∠ACB = (1/2)·∠AOB.
  2. Measure of a central angle = measure of its subtended arc (in degrees).
  3. Angles subtending the same arc are equal: if C and D lie on same arc AB, then ∠ACB = ∠ADB.
  4. Angle in a semicircle: angle subtended by a diameter = 90°.
  5. Opposite angles of a cyclic quadrilateral are supplementary: ∠A + ∠C = 180°, ∠B + ∠D = 180°.
📊 Visual ideas
Basic diagram to illustrate theorem: Draw circle centre O (0,0), radius r. Mark A at angle 0°: A(r,0), B at 60°: B(r·cos60°, r·sin60°). Choose C on the circle at 140°: C(r·cos140°, r·sin140°). Draw radii OA, OB and chord AB; draw lines CA and CB. Label ∠AOB (central) and ∠ACB (inscribed) and measure to verify ∠ACB = 1/2·∠AOB.
Angles subtending same arc: On same circle, pick A and B fixed. Place two points C and D on the same arc AB (not containing the other arc) and draw ∠ACB and ∠ADB to show they are equal. Use coordinates with A at 0°, B at 80°, C at 40°, D at 50° for plotting.
Angle in a semicircle (Thales): Draw diameter endpoints A(-r,0) and B(r,0) and any point C on the semicircle y>0, e.g., C(0,r). Draw triangle ABC to show ∠ACB = 90°.
Cyclic quadrilateral: Choose four points at angles 10°, 110°, 190°, 290° on a circle of radius r. Connect them to form quadrilateral ABCD. Measure opposite angles to demonstrate ∠A + ∠C = 180° and ∠B + ∠D = 180°.
🔢4

Chords: properties and theorems

What is a chord? A chord of a circle is a line segment whose endpoints lie on the circle. A diameter is a chord passing through the centre and is the longest possible chord.

Basic definitions

  • Centre: the fixed point equidistant from all points on the circle.
  • Radius (r): distance from centre to any point on circle.
  • Chord: segment joining two points on the circle.
  • Perpendicular distance (d) from centre to a chord: shortest distance (a perpendicular).

Key properties and theorems (with short justifications)

  • Perpendicular from centre to a chord bisects the chord. If O is centre and AB a chord, drop OM perpendicular to AB. Triangles OMA and OMB are congruent (OA = OB = r, OM common, right angles), so AM = MB.
  • Line through centre that bisects a chord is perpendicular to it (converse). If a line through O bisects AB at M, then OA = OB and OM is common; using triangles again you get right angles at M.
  • Equal chords subtend equal central angles. If chords AB and CD are equal, triangles OAB and OCD (O = centre) are congruent so ∠AOB = ∠COD.
  • Equal chords are equidistant from the centre (and converse). If AB = CD then the perpendicular distances from O to AB and CD are equal. Conversely, chords at equal distances from O are equal in length.
  • Diameter perpendicular to a chord bisects the chord and its arcs. This is a special case of the first property where the line through O is a diameter; it bisects the chord and the two arcs cut by the chord.

Relation between chord length, radius and distance from centre

For a chord of length L at distance d from the centre in a circle of radius r, use the right triangle made by half the chord: (L/2)^2 + d^2 = r^2. Hence

L = 2 sqrt(r^2 - d^2)

Why these matter

These theorems let you construct circles and chords, solve for lengths and angles, and reason about symmetry. They are widely used in geometric constructions, proofs and problem solving.

📌 Examples
  • Bicycle wheel: spokes from the hub to the rim are radii; the straight segment between two rim points is a chord. A spoke perpendicular to a chord (rim segment) would bisect that chord.
  • Design of circular tables or round windows: understand how equally spaced decorative inlays (equal chords) are at the same distance from the centre.
  • Musical instruments (drums): tension lines or chords across the drum head—symmetry about the centre ensures even vibrations when chords are equal or bisected.
  • Bridge arches and circular segments: when constructing supports you often need the midpoint and length of an arch segment, found using perpendicular from the centre.
🧮 Formulas
  1. Definition: chord = segment joining two points on a circle
  2. Diameter is longest chord: diameter = 2r
  3. Perpendicular from centre bisects chord (no formula but a geometric fact)
  4. Chord length in terms of radius r and distance d from centre: L = 2 * sqrt(r^2 - d^2)
  5. Half-chord relation: (L/2)^2 + d^2 = r^2
  6. Equal chords subtend equal central angles: if AB = CD then ∠AOB = ∠COD
📊 Visual ideas
Diagram 1: A circle with centre O, chord AB, perpendicular OM from O to AB. Label OA = OB = r, OM = d, AM = MB = L/2. Use this to show (L/2)^2 + d^2 = r^2.
Diagram 2: Two equal chords AB and CD on opposite sides of O, with perpendiculars OM1 and OM2 showing OM1 = OM2. Label equal lengths to illustrate 'equal chords are equidistant from centre'.
Diagram 3: A diameter through O perpendicular to a chord, showing it bisects the chord and the arcs. Mark arcs with equal tick marks.
Graph 4 (plot): Plot chord length L = 2 * sqrt(r^2 - d^2) versus d for 0 ≤ d ≤ r. This shows L is maximum (2r) when d = 0 and decreases to 0 as d → r.
⭕5

Angles in semicircle and angles in the same segment

Overview: These are basic theorems about angles made by chords and arcs in a circle. They help in identifying right angles and equal inscribed angles.

1. Angle in a Semicircle (Thales' Theorem): If AB is a diameter of a circle and C is any point on the circle (other than A or B), then ∠ACB = 90° (i.e., the angle subtended by a diameter at the circumference is a right angle).

Short proof (coordinate/vector): Put the circle center at the origin with radius r. Let A(−r,0), B(r,0) and C(x,y) such that x² + y² = r². Vectors CA = (−r−x, −y) and CB = (r−x, −y). Their dot product is 0, so CA ∘ CB = 0 ⇒ ∠ACB = 90°.

2. Angles in the Same Segment: If chord AB subtends angles ∠ACB and ∠ADB at two points C and D on the same arc (same side of AB), then ∠ACB = ∠ADB. In words: angles in the same segment of a circle are equal.

Reason (using center O): Let O be the centre. The central angle AOB subtending chord AB equals twice any angle subtended by AB at the circumference: ∠AOB = 2×∠ACB and ∠AOB = 2×∠ADB. Hence ∠ACB = ∠ADB.

Related facts (useful):

  • Angle subtended by a chord at the center = 2 × angle subtended by the same chord at circumference on the same side.
  • Angles subtended by the same chord on the same side are equal.
  • Opposite angles of a cyclic quadrilateral are supplementary (sum to 180°).

How to apply: To prove a triangle is right-angled, show one side is a diameter. To prove two inscribed angles equal, show they subtend the same chord and lie on the same side of it.

📌 Examples
  • Example 1 (Semicircle): A and B are endpoints of diameter of a circle. Point C is on the semicircle. Prove triangle ABC is right-angled at C. (Answer: By Thales' theorem ∠ACB = 90°.)
  • Example 2 (Same segment): In a circle, chord AB subtends ∠ACB = 30° at point C on one arc. Find ∠ADB for any point D on the same arc. (Answer: ∠ADB = 30° since angles in the same segment are equal.)
  • Example 3 (Application): To construct a right angle on the ground: mark two points A and B and draw their circle with AB as diameter; any point C on the circle gives a right angle at C. This is used in field surveying and basic carpentry to produce right angles.
🧮 Formulas
  1. Angle in a semicircle: If AB is diameter, then ∠ACB = 90°.
  2. Angle at centre vs. angle at circumference: ∠AOB = 2 × ∠ACB (where O is centre and C is on the circle).
  3. Angles in the same segment: If points C and D lie on the same arc subtended by chord AB then ∠ACB = ∠ADB.
  4. Cyclic quadrilateral (related): If ABCD is cyclic, then ∠A + ∠C = 180° and ∠B + ∠D = 180°.
📊 Visual ideas
Diagram 1 (Semicircle/Right angle): Draw a circle with centre O, mark diameter endpoints A and B horizontally. Choose a point C on the semicircle above AB. Mark ∠ACB with a right-angle square. Optionally add coordinates A(−r,0), B(r,0), C(x,y) and annotate x² + y² = r²; show dot-product calculation CA·CB = 0.
Diagram 2 (Angles in same segment): Draw circle, chord AB, two points C and D on the same arc above AB. Join A, B, C and A, B, D. Mark ∠ACB and ∠ADB with identical arc markings to show they are equal. Also draw centre O and central angle ∠AOB to indicate ∠AOB = 2×∠ACB = 2×∠ADB.
Interactive suggestion (GeoGebra/Desmos): Create a circle, fix points A and B, make a movable point C on the circle. Display measure of ∠ACB to see it stays 90° when AB is diameter. For same-segment demonstration, fix chord AB and place two movable points C and D on the same arc; show their inscribed angles remain equal as you drag them.
Coordinate sketch: Use circle x² + y² = r² with A(−r,0), B(r,0). Plot several points C(x,y) on the upper semicircle and use vectors CA and CB to visualize perpendicularity (dot product zero).
🔢6

Tangents: definitions and properties

Definition: A tangent to a circle is a straight line that touches the circle at exactly one point. The point where the tangent touches the circle is called the point of contact or point of tangency.

Basic ideas:

  • A tangent meets the circle in exactly one point.
  • A line that intersects a circle in two distinct points is a secant; one that does not meet it at all is an external line.

Important properties (with short justifications):

  • Radius–tangent perpendicularity: The radius drawn to the point of contact is perpendicular to the tangent.
    Justification (idea): If the radius to the contact point were not perpendicular, the line through that radius point would cut the circle at a second point, contradicting that the tangent meets the circle only once. Hence radius at point of contact is perpendicular to the tangent.
  • Converse: A line perpendicular to a radius at its end point on the circle is a tangent.
    Idea: If a line through a point on the circle is perpendicular to the radius, it cannot meet the circle at any other point, so it is a tangent.
  • Equal tangents from an external point: If two tangents are drawn to a circle from an external point P meeting the circle at A and B, then PA = PB.
    Justification (idea): OA and OB are radii. Triangles OAP and OBP are right triangles (OA and OB perpendicular to tangents). Using the equal hypotenuse/leg relationships or Pythagoras with OP common, we get PA = PB.
  • Tangent length formula: If OP = d is the distance from the circle center O to an external point P and r is the radius, then length of the tangent from P to the circle is sqrt(d^2 - r^2).
    Derivation: In right triangle OAP, PA^2 = OP^2 - OA^2 = d^2 - r^2, so PA = sqrt(d^2 - r^2).
  • Angle between tangent and chord (Alternate segment theorem): The angle between a tangent at point A and a chord AB through A equals the angle in the opposite arc (the angle in the alternate segment).
    Idea: This relates a tangent-chord angle to an inscribed angle subtending the same arc.
  • Unique tangent at non-singular point: At any given point on a circle there is exactly one tangent line (except at points where the curve is not smooth, which does not occur for circles).

Use and interpretation: A tangent line gives the instantaneous straight-line direction that just touches the circle without cutting it. In many applications the tangent models the immediate contact direction (for example, a wheel on the road).

Short examples of reasoning you will use:

  • To show two tangent segments from the same external point are equal, form right triangles with the center and apply Pythagoras.
  • To construct a tangent from an external point, draw the line from the external point to the center, draw a circle with that as radius equal to the distance to the center, or use perpendicular from center to the intended tangent point.
📌 Examples
  • A bicycle wheel touches the road at a single point. The road (locally straight) is a tangent to the circular rim at the contact point. The radius drawn to the contact point is perpendicular to the road.
  • From a light pole outside a circular fountain, two ropes are tied tangent to the fountain at A and B. Lengths of the two ropes from the pole to A and to B are equal (equal tangents property).
  • Given a circle of radius 5 cm and a point P at distance 13 cm from the center, length of tangent from P is sqrt(13^2 - 5^2) = sqrt(169 - 25) = sqrt(144) = 12 cm.
  • In optics, the line of sight just grazing a spherical object is approximately a tangent; the tangent-chord angle idea is used in designing reflectors and mirrors.
🧮 Formulas
  1. If O is center, and tangent touches circle at T, then OT ⟂ tangent.
  2. If two tangents from external point P touch circle at A and B then PA = PB.
  3. Tangent length: if OP = d and radius = r, tangent length PA = sqrt(d^2 - r^2).
  4. Equation of tangent (circle x^2 + y^2 = r^2) at point (x1,y1) on circle: x x1 + y y1 = r^2. (coordinate form)
  5. Alternate segment theorem: angle between tangent and chord = angle in opposite arc subtended by the chord.
📊 Visual ideas
Draw a circle with center O and a line touching it at T. Mark OT and show a right angle symbol where OT meets the tangent to illustrate radius ⟂ tangent.
Draw circle with center O and an external point P. Draw two tangents from P touching at A and B. Label PA and PB equal and show right triangles OAP and OBP to illustrate AP = BP and tangent length formula.
Show a chord AB of a circle and the tangent at A. Mark the angle between tangent and chord AB and the inscribed angle subtending arc AB on the opposite side to illustrate the alternate segment theorem.
Coordinate sketch: circle x^2 + y^2 = r^2 and the tangent at point (x1,y1) with line equation x x1 + y y1 = r^2; plot to show intersection at the single point (x1,y1).
🔢7

External tangents and lengths

Definition. A tangent to a circle is a line that touches the circle at exactly one point. If the point from which a tangent is drawn lies outside the circle, that line segment from the external point to the point of contact is called a tangent segment (an external tangent segment).

Key results (Class 9 level).

  • From an external point P to a circle with centre O and radius r, two tangents PA and PB can be drawn. These tangent segments are equal: PA = PB.
  • If OP is the distance from the centre O to the external point P, then the square of the length of the tangent equals the difference of squares of OP and r (Pythagorean relation):
    PA^2 = OP^2 − r^2.

Why this is true (proof idea). Join O to A (point of contact). OA = r and OA ⟂ PA (radius to point of contact is perpendicular to tangent). In triangle OAP, angle at A is 90°, so by Pythagoras OP^2 = OA^2 + AP^2, i.e. AP^2 = OP^2 − r^2. For two tangents from P, triangles OAP and OBP are congruent, giving PA = PB.

Common tangents to two circles. For two circles with centres C1, C2, radii r1, r2 and centre distance d = C1C2, there are two types of common tangents:

  • Direct (external) common tangents — the tangent line touches both circles on the same side of the line joining the centres. The length of the segment between the two points of contact on the tangent is
    L_direct = sqrt(d^2 − (r1 − r2)^2),
    provided the expression under the square root is non‑negative (i.e. the circles are not one strictly inside the other).
  • Transverse (internal) common tangents — the tangent line crosses between the circles; the length of the segment between points of contact is
    L_transverse = sqrt(d^2 − (r1 + r2)^2),
    which exists only when d ≥ r1 + r2 (circles separate enough).

Geometric idea for these formulas. Draw radii to the points of contact; each radius is perpendicular to the tangent. This gives right triangles whose legs include the radii and half (or the whole) of the tangent-segment length; apply Pythagoras to relate d, r1, r2 and the tangent length.

Conditions (quick summary).

  • If d > r1 + r2: the circles are separate — there are 4 common tangents (2 direct, 2 transverse).
  • If d = r1 + r2: circles touch externally — tangent lengths reduce accordingly (one transverse tangent length becomes 0).
  • If |r1 − r2| < d < r1 + r2: the circles intersect — only 2 direct tangents exist.
  • If d = |r1 − r2|: circles touch internally — exactly 1 common tangent.
  • If d < |r1 − r2|: one circle lies inside the other — no common tangents.

Applications: These results are used in construction problems, in optimizing mechanical contacts (pulleys, gears), in optics (tangent rays), and in many 2D geometry problems.

📌 Examples
  • Example 1 — Tangent from an external point: Centre O, radius r = 5 cm. External point P is at distance OP = 13 cm. Length of tangent PA = sqrt(OP^2 − r^2) = sqrt(13^2 − 5^2) = sqrt(169 − 25) = sqrt(144) = 12 cm.
  • Example 2 — Two circles: r1 = 5, r2 = 3, centres distance d = 13. Direct (external) common tangent length L_direct = sqrt(d^2 − (r1 − r2)^2) = sqrt(169 − 4) = sqrt(165) ≈ 12.845. Transverse (internal) common tangent length L_transverse = sqrt(d^2 − (r1 + r2)^2) = sqrt(169 − 64) = sqrt(105) ≈ 10.247.
  • Example 3 — Touching circles: If r1 = 4, r2 = 3 and d = r1 + r2 = 7, then L_transverse = sqrt(d^2 − (r1 + r2)^2) = 0 (they touch externally), while direct tangents still have positive length sqrt(d^2 − (r1 − r2)^2) = sqrt(49 − 1) = sqrt(48).
🧮 Formulas
  1. If P is external point, circle centre O, radius r, and tangent from P meets circle at A: PA^2 = OP^2 − r^2
  2. Equal tangents from same external point: PA = PB
  3. For two circles with centres separated by d and radii r1, r2: length of direct (external) common tangent between contact points: L_direct = sqrt(d^2 − (r1 − r2)^2)
  4. For two circles: length of transverse (internal) common tangent between contact points: L_transverse = sqrt(d^2 − (r1 + r2)^2)
  5. Existence conditions: L_direct real ⇔ d ≥ |r1 − r2|; L_transverse real ⇔ d ≥ r1 + r2
📊 Visual ideas
Diagram A (single circle, external point): Draw circle centre O, radius r. Mark external point P at distance OP. Draw two tangents from P meeting circle at A and B. Show OA and OB perpendicular to tangents and triangle OAP with right angle at A — label OP, OA = r, PA and apply Pythagoras.
Diagram B (two circles on same baseline): Place centres C1 at (0,0) and C2 at (d,0). Draw circles of radii r1 and r2. Draw one direct common tangent above the x-axis touching circles at T1 and T2. Draw radii C1T1 and C2T2 perpendicular to tangent and the right triangles used to derive L_direct; annotate d, r1, r2 and L_direct.
Diagram C (internal/transverse tangent): Same centre placement. Draw the transverse tangent that crosses between circles and touches at S1, S2. Show perpendicular radii and the right-triangle relation giving L_transverse.
Coordinate sketch suggestion: Put circle1 centre (0,0), circle2 centre (d,0). For a tangent with slope m, write line y = m x + c and solve system with circle equations to find tangency condition (discriminant = 0). Use this to illustrate algebraic derivation and to plot tangent lines numerically.
🔢8

Arcs, sectors and segments (definitions and relations)

Definitions

Circle: set of all points in a plane at a fixed distance (radius r) from a fixed point (center O).

Arc: a continuous part of the circumference between two points A and B. The shorter arc between A and B is the minor arc; the longer is the major arc. An arc is usually specified by its central angle θ (angle AOB).

Sector: the region enclosed by two radii OA and OB and the arc AB. The region with the smaller central angle is the minor sector, the other is the major sector. A sector is like a "slice" of the circle.

Segment: the region bounded by a chord AB and the arc AB. The region bounded by the chord and the smaller arc is the minor segment; with the larger arc is the major segment. A segment = sector minus the isosceles triangle formed by the two radii.

Relations and key ideas

  • The central angle θ (in degrees or radians) determines arc length and sector area: arc length is proportional to θ; sector area is proportional to θ.
  • Radians provide the simplest formulas: 1 full circle = 2π radians; to convert degrees to radians multiply by π/180 (θ_rad = θ_deg * π/180).
  • Chord length is related to central angle: c = 2r sin(θ/2) (θ in radians or degrees as long as sine uses same-angle unit).
  • Area of the isosceles triangle OAB formed by the two radii = (1/2) r^2 sin θ (θ in radians or degrees—using sine accordingly).
  • Area of a segment = area of corresponding sector − area of triangle OAB. For the major segment, subtract the minor segment from the whole circle area.
  • Perimeter of a sector = arc length + two radii = (arc length) + 2r. Perimeter of a segment = arc length + chord length.

Units and tips

  • Keep angle units consistent when using formulas (radians simplify many formulas).
  • When θ is given in degrees, use the fraction θ/360 for parts of the full circle.
📌 Examples
  • Real-life examples: a pizza slice (sector), a slice of orange between membranes (segment), the curved edge between two points on a running track (arc), sections of a circular cake (sectors), and the water area cut off by a retaining wall in a circular pond (segment).
  • Numeric example: Radius r = 7 cm, central angle θ = 60°. Convert to radians: θ = 60° = π/3 rad. - Arc length s = rθ = 7 * (π/3) = 7π/3 ≈ 7.33 cm. - Sector area = (1/2) r^2 θ = 0.5 * 49 * (π/3) = 49π/6 ≈ 25.67 cm². - Triangle area (OAB) = (1/2) r^2 sin θ = 0.5 * 49 * sin 60° = 24.5 * (√3/2) ≈ 21.22 cm². - Segment area = sector area − triangle area ≈ 25.67 − 21.22 = 4.45 cm².
  • Using chord length: For the same example, chord c = 2r sin(θ/2) = 2*7*sin(30°) = 14 * 0.5 = 7 cm. Perimeter of the minor sector = arc + 2r ≈ 7.33 + 14 = 21.33 cm. Perimeter of the minor segment = arc + chord ≈ 7.33 + 7 = 14.33 cm.
🧮 Formulas
  1. Conversion: θ (radians) = θ (degrees) * π/180, and θ (degrees) = θ (radians) * 180/π.
  2. Arc length (s): s = rθ (θ in radians) OR s = (θ_deg/360) * 2πr.
  3. Sector area (A_sector): A = (1/2) r^2 θ (θ in radians) OR A = (θ_deg/360) * πr^2.
  4. Chord length (c): c = 2r sin(θ/2).
  5. Area of triangle OAB: A_triangle = (1/2) r^2 sin θ.
  6. Segment area (minor): A_segment = A_sector − A_triangle = (1/2) r^2 (θ − sin θ) (θ in radians).
📊 Visual ideas
Diagram 1 (basic labels): Draw a circle with center O, points A and B on circumference. Draw radii OA and OB. Mark central angle ∠AOB = θ. Shade the minor arc AB. Label radius r, arc length s, chord AB.
Diagram 2 (sector): Same circle; shade the region bounded by OA, OB and arc AB (the sector). Label sector area and show formula A = (1/2) r^2 θ. Annotate sector perimeter = rθ + 2r.
Diagram 3 (segment): Draw chord AB and shade the smaller region between chord AB and arc AB (minor segment). Also draw triangle OAB. Show that segment area = sector area − area of triangle. Mark triangle height and chord length c = 2r sin(θ/2).
Interactive graph suggestion: Use a circle centred at origin with radius r. Let A and B be points at angles +θ/2 and −θ/2. Provide sliders for r and θ to dynamically show changes in arc length s = rθ, sector area, chord length, triangle area and segment area. Use contrasting colors for arc, sector and segment; annotate numeric values as sliders move.
⭕9

Constructions related to circles

Overview: "Constructions related to circles" covers methods to draw tangents and circles under given conditions using ruler and compass, and the geometric reasoning behind them. Key ideas used are: the radius to a point of contact is perpendicular to the tangent, and Thales' theorem (angle in a semicircle is a right angle).

Basic constructions and reasoning:

  1. To draw a circle when the centre and radius are given — Place the compass at the centre, open it to the given radius and draw the circle. (This is the basic starting construction.)
  2. Tangent to a circle at a given point on the circle
    1. Given circle with centre O and point A on the circle. Join O to A.
    2. At A, draw a line perpendicular to OA. That line is the tangent to the circle at A.
    3. Reason: OA is a radius and the radius at the point of contact is perpendicular to the tangent.
  3. Tangents from an external point to a circle (construct both tangents)
    1. Let O be the centre of the circle and P be a point outside the circle.
    2. Join OP. Construct the circle having OP as diameter (i.e., with centre M, the midpoint of OP, and radius = OP/2).
    3. The circle with diameter OP intersects the given circle at the points of contact T₁ and T₂ (if any). Join P to each intersection; PT₁ and PT₂ are the required tangents.
    4. Reason: For a tangent PT, angle OTP is 90°. By Thales' theorem any point T on the circle with diameter OP satisfies ∠OTP = 90°, so intersection points give right angles and thus tangency.
  4. Circle through three non-collinear points (circumcircle)
    1. Given three non-collinear points A, B, C. Draw segments AB and BC.
    2. Construct perpendicular bisectors of AB and BC. Their intersection O is the centre of the circle passing through A, B and C.
    3. With centre O and radius OA draw the circle — it passes through A, B and C.
    4. Reason: Points equidistant from A and B lie on the perpendicular bisector of AB. Intersection of two such bisectors gives a point equidistant from A, B and C — the centre.
  5. Length of tangent from an external point (useful relation)

    If P is outside the circle with centre O and radius r, and PT is the tangent length from P to the point of contact T, then PT = sqrt(OP^2 − r^2). This follows from right triangle OTP where OT = r and ∠OT P = 90°.

Practical tips when constructing:

  • Always mark the centre and given points clearly and use a sharp pencil for compass arcs.
  • To find the midpoint of a segment, draw equal arcs from endpoints and join their intersection points — then draw the perpendicular bisector.
  • When constructing tangents from P, if OP < r then no real tangent exists (point is inside the circle); if OP = r exactly one tangent exists (point on the circle).

Connections to proofs and problems: These constructions are used to prove properties (radius ⟂ tangent, equal tangents from an external point) and to solve geometric construction problems such as drawing a circle that touches given lines or passes through given points.

📌 Examples
  • Automobile wheel on a road: the road is locally tangent to the circular tyre at the point of contact. Drawing the tangent line at the contact point helps analyze forces and friction.
  • Designing a roundtable that must pass through three existing pillars: construct the circumcircle through the three pillar positions to find the table's centre and radius.
  • Drawing tangents from an external point: connecting a lamp post (external point) to a circular fountain so that support ropes just touch the fountain — use tangent construction to find contact points and rope lengths.
  • Engineering gears and cams: the instantaneous direction of motion at the contact point is tangent to the circular profile; tangent construction helps visualize contact direction.
🧮 Formulas
  1. Radius ⟂ Tangent: If OT is radius and PT is tangent at T, then OT ⟂ PT.
  2. Length of tangent from external point P: PT = sqrt(OP^2 − r^2), where O is centre and r is radius (provided OP ≥ r).
  3. Condition for tangency from P: Number of real tangents = 0 if OP < r, 1 if OP = r (point on circle), 2 if OP > r.
📊 Visual ideas
Tangent at a point: Plot circle centre O at (0,0) with radius 3. Take point A at (3,0). Draw radius OA and the tangent line x=3 (vertical). Label OA and show the right angle at A.
Tangents from an external point: Plot circle centre O at (0,0) with radius 3 and external point P at (5,0). Draw line OP, midpoint M at (2.5,0). Draw circle with diameter OP (centre M, radius 2.5). Mark intersections T1,T2 between the two circles and draw PT1 and PT2 as tangents. Annotate distances OP, PT and r.
Circumcircle through three points: Plot three non-collinear points A(1,2), B(4,1), C(2,4). Draw AB and BC, construct perpendicular bisectors, mark their intersection O and draw the circumcircle through A,B,C. Show equal radii OA=OB=OC.
Right-triangle relation for tangent length: On the same graph as the second example, draw triangle OTP and show OT = r, OP labeled, and PT as the tangent; annotate PT = sqrt(OP^2 − r^2).
🔢10

Problem types and proof techniques

Overview
In the Class 9 chapter on Circles, many questions ask you to prove relationships between angles, chords, radii and tangents. Typical problems require applying a few basic circle theorems together with elementary triangle congruence and isosceles-triangle arguments.

Common problem types

  • Angle relations: show an angle subtended by a chord at the centre is twice the angle subtended at the circumference, angle in a semicircle is 90°, angles in the same segment are equal.
  • Chord and distance relations: prove a perpendicular from the centre bisects a chord, prove equal chords are equidistant from the centre and vice versa.
  • Tangent problems: show a tangent is perpendicular to the radius at the point of contact, prove lengths of tangents from an external point are equal, relate tangent–chord angle to angle in the opposite arc.
  • Proofs combining the above: e.g. prove a quadrilateral is cyclic or prove right angles using chord/arc properties.

Proof techniques and strategy

  • Draw a clear diagram and label given elements (centre O, chord AB, point P etc.).
  • Mark equal lengths and parallel/perpendicular lines as you infer them.
  • Use congruent triangles (SSS, SAS, ASA, RHS) to prove equality of angles or lengths. Many circle proofs reduce to showing two triangles are congruent.
  • Exploit isosceles triangles: if OA = OB (radii), triangle OAB is isosceles so base angles are equal.
  • Apply basic circle theorems: central angle = 2 × inscribed angle; angle in a semicircle = 90°; perpendicular from centre bisects chord; equal chords subtend equal angles; tangent ⟂ radius at contact.
  • If direct approach is messy, use proof by contradiction: assume the negation and derive an impossible result.
  • Write a short, logical sequence: what is given → which theorem/triangle used → conclusion.

Tips

  • Always refer to points precisely (e.g. ∠AOB, chord AB, arc ACB) so your reasoning is unambiguous.
  • When you claim triangles are congruent, explicitly state which congruence criterion you use and which corresponding parts match.
  • Practice sketching accurate diagrams — many mistakes come from incorrect or incomplete figures.

📌 Examples
  • Problem 1: Prove that the angle in a semicircle is a right angle. Sketch and solution: Let AB be a diameter of circle with centre O and C any point on the circle. In triangles OCA and OCB, OA = OC = OB = radius. Show triangles OCA and OCB are isosceles and use angle sum in ΔABC to get ∠ACB = 90°.
  • Problem 2: Prove that the perpendicular from the centre of a circle to a chord bisects the chord. Sketch and solution: Let AB be a chord, O the centre, and OD ⟂ AB at D. In triangles OAD and OBD, OA = OB (radii) and OD is common; with right angles, triangles are congruent (RHS), so AD = DB.
  • Problem 3: If equal chords AB and CD are in a circle with centre O, prove OA = OC and show AB and CD are equidistant from O. Sketch and solution: Equal chords subtend equal central angles ∠AOB = ∠COD. Use isosceles triangles to show distances from O to midpoints are equal.
  • Problem 4: Prove the tangent–chord theorem (angle between tangent and chord equals angle in the opposite arc). Sketch and solution: Let PT be a tangent at T and chord TB with point A on arc opposite to TB. Join OT and show triangle relations; using central/inscribed angle facts you get ∠PTB = ∠PAB (angle in alternate segment).
  • Problem 5: Tangent lengths from an external point P to circle with contact points A and B are equal. Sketch and solution: Join PA and PB to centre O. Triangles OAP and OBP have OA = OB (radii) and ∠OAP = ∠OBP = 90°. Use RHS congruence to get PA = PB.
🧮 Formulas
  1. Central–inscribed angle relation: ∠AOB = 2 × ∠ACB (for chord AB and point C on the circle).
  2. Angle in a semicircle: If AB is a diameter then for any C on the circle, ∠ACB = 90°.
  3. Angles in same segment: Angles subtended by the same chord and on the same side of the chord are equal.
  4. Perpendicular from centre to chord: The line from centre O perpendicular to chord AB bisects AB and the corresponding arc.
  5. Equal chords–equal arcs–equal angles: Equal chords subtend equal arcs and equal central angles; conversely equal central angles imply equal chords.
  6. Tangent–radius: A tangent at point T is perpendicular to radius OT (OT ⟂ tangent at T).
📊 Visual ideas
Diagram 1: Circle with centre O, diameter AB, point C on circle. Show angle ∠ACB (inscribed) and ∠AOB (central) to illustrate central–inscribed relation.
Diagram 2: Circle with centre O and chord AB. Draw OD ⟂ AB meeting at D. Label OA and OB. Use this to show OD bisects AB.
Diagram 3: Circle showing two equal chords AB and CD on either side of O. Draw perpendiculars from O to the chords to show equal distances.
Diagram 4: Circle with tangent at T, radius OT drawn, and a chord TB. Mark angle between tangent and chord and the inscribed angle subtending the chord on the opposite arc.

Key Concepts

Circle
The set of all points in a plane at a fixed distance (radius) from a fixed point (centre).
Centre
The fixed point from which all points on a circle are equidistant.
Radius
A line segment joining the centre of a circle to any point on the circle; its length is called the radius.
Diameter
A chord that passes through the centre; its length is twice the radius.
Chord
A line segment whose endpoints lie on the circle.
Tangent
A line that touches the circle at exactly one point and does not cross it at that neighbourhood.
Secant
A line that intersects a circle at two distinct points.
Point of contact
The single point where a tangent touches the circle.
Arc
A continuous part of the circumference between two points on the circle.
Minor arc
The smaller arc connecting two points on a circle (measure < 180°).
Major arc
The larger arc connecting two points on a circle (measure > 180°).
Semicircle
An arc (or region) formed by a diameter; its measure is 180°.
Sector
The region bounded by two radii and the included arc (like a pizza slice).
Segment (of a circle)
The region bounded by a chord and the corresponding arc (minor or major segment).
Central angle
An angle whose vertex is the centre of the circle and whose sides are radii.
Inscribed angle (angle subtended at the circumference)
An angle formed by two chords with its vertex on the circle; it subtends an arc.
Circumference
The total distance around a circle; length = 2πr where r is the radius.
Concentric circles
Two or more circles that share the same centre but have different radii.
Equal chords
Chords of a circle that have the same length; equal chords subtend equal arcs and are equidistant from the centre.
Perpendicular bisector of a chord
A line perpendicular to a chord through its midpoint; in a circle it passes through the centre.

End-of-Chapter Trial Paper & Test Questions

Topic-wise questions to test your understanding of every concept in this chapter.

  1. Which of the following is the longest chord of a circle? / निम्नलिखित में से कौन सा वृत्त की सबसे लंबी जीवा है? (a) Radius / त्रिज्या (b) Tangent / स्पर्श रेखा (c) Diameter / व्यास (d) Secant / छेदक
    Show answer

    (c) Diameter / व्यास — The diameter passes through the centre and its length is 2r, which is the maximum possible chord length. / व्यास केंद्र से होकर गुजरता है और इसकी लंबाई 2r होती है, जो अधिकतम संभव जीवा की लंबाई है।

  2. A chord AB of a circle is at a distance of 5 cm from the centre. If the radius is 13 cm, what is the length of the chord? / एक वृत्त की जीवा AB केंद्र से 5 सेमी की दूरी पर है। यदि त्रिज्या 13 सेमी है, तो जीवा की लंबाई क्या है? (a) 8 cm / 8 सेमी (b) 12 cm / 12 सेमी (c) 18 cm / 18 सेमी (d) 24 cm / 24 सेमी
    Show answer

    (d) 24 cm / 24 सेमी — Half-chord = √(r² − d²) = √(169 − 25) = √144 = 12 cm, so full chord = 2 × 12 = 24 cm. / अर्ध-जीवा = √(r² − d²) = √(169 − 25) = 12 सेमी, अतः पूरी जीवा = 24 सेमी।

  3. The angle subtended by a diameter at any point on the circle is: / वृत्त पर किसी भी बिंदु पर व्यास द्वारा अंतरित कोण होता है: (a) 45° (b) 60° (c) 90° (d) 180°
    Show answer

    (c) 90° — By Thales' theorem, the angle in a semicircle is always 90°. / थेल्स प्रमेय के अनुसार, अर्धवृत्त में कोण सदैव 90° होता है।

  4. Fill in the blank: The perpendicular from the centre of a circle to a chord _____ the chord. / रिक्त स्थान भरें: वृत्त के केंद्र से जीवा पर डाला गया लंब जीवा को _____ करता है।
    Show answer

    bisects / समद्विभाजित — The perpendicular from the centre to a chord is a line of symmetry, so it bisects the chord (AM = MB). / केंद्र से जीवा पर लंब एक सममिति रेखा है, अतः यह जीवा को समद्विभाजित करती है।

  5. Fill in the blank: Equal chords of a circle are _____ from the centre. / रिक्त स्थान भरें: एक वृत्त की समान जीवाएँ केंद्र से _____ होती हैं।
    Show answer

    equidistant (equal distance) / समदूरस्थ (समान दूरी पर) — Equal chords subtend equal central angles, and so they are at equal perpendicular distances from the centre. / समान जीवाएँ केंद्र पर समान कोण अंतरित करती हैं और इसलिए केंद्र से समान लंब दूरी पर होती हैं।

  6. True or False: Angles subtended by the same chord at two different points on the same arc of a circle are equal. / सत्य या असत्य: वृत्त के एक ही चाप पर दो भिन्न बिंदुओं पर एक ही जीवा द्वारा अंतरित कोण बराबर होते हैं।
    Show answer

    True / सत्य — This is the theorem of angles in the same segment: they both equal half the central angle subtending the same arc. / यह एक ही खंड में कोणों का प्रमेय है: दोनों कोण उसी चाप पर केंद्रीय कोण के आधे के बराबर होते हैं।

  7. From an external point P, a tangent PT is drawn to a circle with centre O and radius 5 cm. If OP = 13 cm, find the length of the tangent PT. / एक बाह्य बिंदु P से, केंद्र O और त्रिज्या 5 सेमी वाले वृत्त पर एक स्पर्श रेखा PT खींची गई है। यदि OP = 13 सेमी है, तो स्पर्श रेखा PT की लंबाई ज्ञात कीजिए।
    Show answer

    PT = √(OP² − OT²) = √(13² − 5²) = √(169 − 25) = √144 = 12 cm / PT = 12 सेमी। OT is perpendicular to PT (radius to tangent), so Pythagoras applies. / OT, PT पर लंब है (त्रिज्या से स्पर्श रेखा), अतः पाइथागोरस प्रमेय लागू होता है।

  8. What is the relation between a central angle and the inscribed angle subtending the same arc? / एक ही चाप पर केंद्रीय कोण और परिलेखित कोण के बीच क्या संबंध है?
    Show answer

    The central angle is twice the inscribed angle subtending the same arc (∠AOB = 2 × ∠ACB, where C is on the circle). / केंद्रीय कोण, उसी चाप पर परिलेखित कोण का दोगुना होता है (∠AOB = 2 × ∠ACB, जहाँ C वृत्त पर है)। This is the central-inscribed angle theorem. / यह केंद्रीय-परिलेखित कोण प्रमेय है।

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