Overview
This chapter studies the geometry of circles: their basic elements (centre, radius, diameter, chord, arc, sector, segment) and fundamental properties and theorems about chords, arcs and tangents. It develops methods to prove and apply results such as equal chords subtending equal angles at the centre, perpendicular from the centre bisects the chord, angles in the same segment are equal, the angle in a semicircle is a right angle, and key tangent properties (radius ⟂ tangent, equal tangents from an external point, angle between tangent and chord). Importance: these results are central to Euclidean geometry and frequently appear in CBSE questions; they build logical reasoning, diagram-based proof skills and problem-solving techniques. What you will learn: precise definitions, statement and proofs of main theorems, solving numerical and proof-based exercises using these theorems, and how to apply circle properties to construct and analyse geometrical figures.
Learning Objectives
- Define circle, centre, radius, diameter, chord, arc, sector and segment.
- State and prove that equal chords of a circle subtend equal angles at the centre and use this to solve problems.
- Explain and apply the result that the perpendicular from the centre to a chord bisects the chord and the corresponding arcs.
- Prove that the angle in a semicircle is a right angle and solve related numerical questions.
- Apply the theorem that angles in the same segment of a circle are equal to determine unknown angles.
- Determine the relation between an angle subtended at the centre and the angle subtended at the circumference by the same arc and use it in calculations.
- State and apply properties of cyclic quadrilaterals, including that opposite angles are supplementary, to solve problems.
- Define a tangent to a circle and prove that a tangent is perpendicular to the radius at the point of contact.
Topics in this chapter
6 topics · tap a topic title to jump straight to it.
Basic definitions and parts of a circle
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 the center and the fixed distance is the radius.
- Center (O): The fixed point from which every point on the circle is equidistant.
- Radius (r): A line segment joining the center to any point on the circle. All radii are equal.
- Diameter (d): A chord passing through the center. Its length d = 2r.
- Chord: A line segment joining two points on the circle. A diameter is the longest chord.
- Arc: A continuous part of the circle between two points. The shorter arc is the minor arc; the longer is the major arc.
- Central angle: An angle whose vertex is at the center and whose sides cut the circle at two points. The measure of a central angle equals the measure of its intercepted arc (in degrees).
- Sector: The region bounded by two radii and the included arc (like a pizza slice).
- Segment: The region bounded by a chord and the corresponding arc (area of sector minus triangle formed by the two radii).
- Semicircle: A region enclosed by a diameter and the arc it subtends (half the circle).
- Tangent: A line that touches the circle at exactly one point. At the point of contact P, the radius OP is perpendicular to the tangent.
- Secant: A line that cuts the circle at two points.
- Concentric circles: Two or more circles with the same center but different radii.
Useful properties (short):
- The perpendicular from the center to a chord bisects the chord (and the corresponding arcs).
- Equal chords subtend equal arcs and are equidistant from the center.
- The tangent at a point is perpendicular to the radius drawn to the point of contact.
These definitions and properties form the basis for solving many problems on arcs, sectors, chords and tangents in Class 10 mathematics.
- Clock face: the rim is a circle; hour markings lie on the circumference. The hands sweep out sectors and arcs.
- Pizza or cake: a slice is a sector (bounded by two radii and an arc); the crust between two cut points is an arc or segment depending on the cut.
- Wheel or tyre: spokes are radii, the rim is the circle, diameter is the distance across the wheel through the center.
- Coin and a straight edge: the line touching the coin at exactly one point is a tangent; a line cutting through the coin is a secant.
- Circular pond with concentric paths: paths having same center but different widths are concentric circles.
- Semicircular bridge or half-round arch: the arch shape is a semicircle—useful in area and perimeter calculations.
- Diameter: d = 2r
- Circumference (perimeter) of circle: C = 2πr = πd
- Area of circle: A = πr²
- Length of an arc (central angle θ in degrees): L = (θ/360) × 2πr = (θπ/180) × r
- Length of an arc (angle θ in radians): L = rθ
- Area of a sector (θ in degrees): As = (θ/360) × πr²
Chord properties
Definition: A chord of a circle is a line segment whose endpoints lie on the circle. The line joining the center of the circle to a chord (at right angle) is important in chord geometry.
Key properties (Class 10 level) with short reasoning:
- Perpendicular from centre bisects the chord: If O is the centre and AB is a chord, the perpendicular from O to AB meets AB at M and OM ⟂ AB ⇒ AM = MB. Reason: In ΔOAM and ΔOBM, OA = OB (radii), OM is common and both have a right angle, so triangles are congruent.
- Perpendicular bisector of a chord passes through the centre: Any line perpendicular to a chord at its midpoint passes through the centre. (Conversely of previous property.)
- Equal chords subtend equal angles at the centre: If AB and CD are equal chords, then ∠AOB = ∠COD (O is centre). Reason: In isosceles triangles formed by radii, equal base lengths imply equal vertex angles.
- Chords equidistant from the centre are equal (and vice versa): If distance from O to chord AB equals distance from O to chord CD, then AB = CD. Conversely, equal chords are at equal distances from the centre. Reason: Use right triangles formed with OM (distance) and half-chord lengths.
- Diameter perpendicular to a chord bisects the chord and its arc: If a diameter is perpendicular to a chord, it bisects the chord and the arcs cut off by the chord.
Geometric setup and derived relations: Let AB be a chord, O the centre, M the midpoint of AB, R the radius, c the chord length AB, d = OM (perpendicular distance from centre to chord), and θ = central angle AOB (in radians or degrees). From right triangle OMA we get:
- c/2 = R·sin(θ/2) ⇒ c = 2R·sin(θ/2)
- d = R·cos(θ/2)
- or eliminating θ: c = 2·sqrt(R^2 − d^2) and d = sqrt(R^2 − (c/2)^2)
Uses: These properties are used to solve problems involving symmetry of circles, locating the centre using chords, computing chord lengths from central angles (and vice versa), and designing circular parts in engineering and architecture.
- Bicycle wheel: The rim is a circle and segments of the rim between two spokes are chords. Equal spacing of spokes creates equal chords equidistant from the centre, giving balance.
- Bridge arches or circular windows: Support struts sometimes connect two points on a circular arc; the strut is a chord. If a support (diameter) is perpendicular to a chord, it will bisect the chord — useful for symmetric design.
- Musical instruments (string sections on a circular soundboard): Equal chords correspond to equal angular spans and produce symmetric vibration regions.
- Numerical example: Let radius R = 10 cm and central angle θ = 60°. Chord length c = 2R·sin(θ/2) = 2·10·sin30° = 20·0.5 = 10 cm. The perpendicular distance from centre to chord d = R·cos30° = 10·(√3/2) ≈ 8.66 cm.
- Another numeric: Given R = 13 cm and chord length c = 10 cm, half-chord = 5 so d = sqrt(R^2 − (c/2)^2) = sqrt(169 − 25) = sqrt(144) = 12 cm. The chord lies 12 cm from the centre.
- Chord length from central angle: c = 2R · sin(θ/2) (θ is central angle AOB)
- Distance from centre to chord: d = R · cos(θ/2)
- Chord–distance relation: c = 2 · sqrt(R^2 − d^2)
- Inverse (distance from c): d = sqrt(R^2 − (c/2)^2)
- Arc length (related): s = R · θ (θ in radians) — useful to connect chord and arc
Angles in a circle
Basic idea. An angle is said to be subtended by an arc or a chord at a point if the two rays of the angle pass through the end-points of the arc/chord. In a circle, relationships among angles subtended by the same arc or chord (at the centre, at the circumference, by tangents, and in cyclic quadrilaterals) are important and frequently used.
Key theorems (with brief reasons):
- Angle at the centre and angle at the circumference: An angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at any point on the remaining part of the circumference. (If arc AB subtends ∠AOB at centre O and ∠ACB at point C on circumference, then ∠AOB = 2∠ACB.) Reason: join O to A, B and use isosceles triangles OAC, OBC to relate base angles.
- Angles in the same segment: Angles subtended by the same chord and on the same side of the chord are equal. (If chord AB subtends ∠APB and ∠AQB at two points P and Q on the same arc, then ∠APB = ∠AQB.) This follows from the previous theorem by comparing both angles to the central angle subtending AB.
- Angle in a semicircle: An angle subtended by a diameter is a right angle. (If AC is a diameter and B is any other point on the circle, then ∠ABC = 90°.) The converse is also true: if an angle subtended by a chord is 90°, that chord is a diameter.
- Cyclic quadrilateral (opposite angles): If four points A, B, C, D lie on a circle (a cyclic quadrilateral), then opposite angles are supplementary: ∠A + ∠C = 180° and ∠B + ∠D = 180°. This follows by writing each angle as an angle subtended by some arc and using the centre–circumference relation.
- Tangent–chord theorem: The angle between a tangent and a chord through the point of contact equals the angle in the opposite arc (the angle subtended by the chord at the far side of the circle). If PT is a tangent at T and chord TA meets the circle at A, then ∠PTA = angle in the arc opposite TA.
How to apply these theorems. Typical problems ask you to find unknown angles using combinations of the above facts: convert central angles to circumference angles, use equality of angles in the same segment, use the right-angle property for diameters, and use supplementary relationships in cyclic quadrilaterals and tangent–chord equalities.
Short worked idea: If central angle ∠AOB = 140°, then any angle subtended by arc AB at the circumference = 140°/2 = 70°. If a diameter BD subtends ∠BAD then ∠BAD = 90°.
- Numeric: In circle with centre O, if central angle ∠AOB = 120°, then any angle ∠ACB subtended by arc AB at the circumference = 60° (because ∠AOB = 2∠ACB).
- Cyclic quadrilateral: If ABCD is cyclic and ∠A = 110°, then ∠C = 70° because opposite angles are supplementary (110° + 70° = 180°).
- Semicircle: For a circle with diameter AB, any point C on the circle (not A or B) gives ∠ACB = 90°. Example: if AB is a diameter of a wheel, the triangle formed by two rim points and the far end of the diameter is right-angled.
- Tangent–chord: If PT is tangent at T and chord TA subtends angle ∠TBA = 40° at a point B on the opposite arc, then ∠PTA = 40° (angle between tangent and chord equals angle in opposite arc).
- Real-life: Clock face — the angle between two hands as they move can be analyzed using central angles; sundials — shadow angles correspond to angles subtended at the centre of the circular dial; design of circular windows and arches uses equal angles in segments for symmetry.
- Angle subtended at centre = 2 × angle subtended at circumference by same arc (∠AOB = 2∠ACB).
- Angles in the same segment are equal (∠APB = ∠AQB for points P,Q on same arc AB).
- Angle in a semicircle = 90° (angle subtended by a diameter).
- Opposite angles of a cyclic quadrilateral are supplementary (∠A + ∠C = 180°, ∠B + ∠D = 180°).
- Angle between tangent and chord = angle in the opposite arc.
Cyclic quadrilateral
Definition: A cyclic quadrilateral is a four-sided polygon whose four vertices all lie on the circumference of the same circle.
Key properties and short proofs:
- Opposite angles are supplementary: If ABCD is cyclic (vertices in order), then ∠A + ∠C = 180° and ∠B + ∠D = 180°. Proof idea: an inscribed angle equals half the measure of its intercepted arc. The arcs intercepted by a pair of opposite angles together make the full circle (360°), so their half-sum is 180°.
- Converse: If a quadrilateral has a pair of opposite angles supplementary, then it is cyclic. Proof idea: construct the circumcircle of triangle formed by three vertices; the supplementary condition forces the fourth vertex to lie on that circle.
- Exterior angle property: An exterior angle of a cyclic quadrilateral equals the interior opposite angle. For example, exterior angle at A equals ∠C.
- Ptolemy's theorem: For a cyclic quadrilateral ABCD, the product of the diagonals equals the sum of the products of opposite sides: AC·BD = AB·CD + BC·AD.
- Area (Brahmagupta's formula): If a, b, c, d are the side lengths and s = (a+b+c+d)/2, area = sqrt((s-a)(s-b)(s-c)(s-d)). (This holds only for cyclic quadrilaterals.)
Special cases: Rectangles, squares and isosceles trapeziums are cyclic. A general kite is not necessarily cyclic.
Uses and significance: The cyclic property simplifies angle-chasing and length relations in geometry problems and appears in many constructions, proofs and geometry-based design tasks.
- Architectural rose windows or stained-glass window panels where decorative quadrilaterals are inscribed in circular frames (each corner touches the circular frame).
- A clock face: pick four hour marks on the circular dial and connect them — the quadrilateral formed by the four hour points is cyclic.
- Surveying and triangulation sketches: when four fixed landmark points lie approximately on a circle, the quadrilateral they form can be treated as cyclic for angle and distance calculations.
- Designs in jewellery (pendants, medallions) where a decorative quadrilateral is inscribed inside a circular bezel.
- Opposite angles: ∠A + ∠C = 180°, ∠B + ∠D = 180°
- Exterior angle: Exterior angle at a vertex = interior opposite angle (e.g., exterior at A = ∠C)
- Ptolemy's theorem: AC × BD = AB × CD + BC × AD
- Brahmagupta area: Area = sqrt((s-a)(s-b)(s-c)(s-d)), where s = (a+b+c+d)/2
- Inscribed angle relation: An inscribed angle = 1/2 × measure of its intercepted arc (useful to prove angle properties)
Tangent to a circle
Definition: A tangent to a circle is a straight line that touches the circle at exactly one point. That point is called the point of contact.
Basic properties (with short reasoning):
- Radius to point of contact is perpendicular to the tangent: If O is the centre and P the point of contact, then OP ⟂ tangent. (If the line through P were not perpendicular to OP it would meet the circle at a second point, contradicting tangency.)
- From an external point A, you can draw two tangents to a circle; the tangent segments from A to the circle are equal in length. (Use right triangles formed by radii to points of contact.)
- A line is tangent to a circle iff the distance from the circle's centre to the line equals the radius.
Coordinate-geometry viewpoint (short): For circle (x − a)^2 + (y − b)^2 = r^2, the tangent at a point (x1, y1) on the circle has equation (x1 − a)(x − a) + (y1 − b)(y − b) = r^2. For the circle x^2 + y^2 = r^2 this simplifies to x x1 + y y1 = r^2. A line y = m x + c is tangent to x^2 + y^2 = r^2 exactly when c^2 = r^2(1 + m^2).
Applications / Intuition: Tangents give the instantaneous direction of a circle's boundary; they are used to approximate the circle locally by a straight line, to compute shortest contact distances, and are fundamental in problems involving angles, optics, gears and motion.
- Wheel on a road: the ground is tangent to the circular tyre at the contact point. The radius to the contact point is perpendicular to the road there.
- Touching circles / gears: teeth contact along tangents; the point of contact lies on the common tangent where the radii to the contact point are perpendicular to that tangent.
- Tangent plane to Earth approximated locally: for small regions, the Earth’s surface is approximated by a tangent plane (centre at Earth’s centre, radius ~ Earth’s radius).
- From an observer outside a circular pond, two sight-lines that just skim the pond are tangents; the visible tangent segments from the observer are equal in length.
- Condition for line lx + my + n = 0 to be tangent to circle (x − a)^2 + (y − b)^2 = r^2: |l a + m b + n| / √(l^2 + m^2) = r.
- Equation of tangent at (x1, y1) on circle (x − a)^2 + (y − b)^2 = r^2: (x1 − a)(x − a) + (y1 − b)(y − b) = r^2.
- For circle x^2 + y^2 = r^2, tangent at (x1, y1): x x1 + y y1 = r^2.
- Slope form: line y = m x + c is tangent to x^2 + y^2 = r^2 iff c^2 = r^2(1 + m^2).
- Length of tangent from external point (x1, y1) to circle (x − a)^2 + (y − b)^2 = r^2: √[(x1 − a)^2 + (y1 − b)^2 − r^2].
- If two tangents from external point A meet circle at P and Q, then AP = AQ (tangent segments from same external point are equal).
Applications and problem-solving
This topic applies the main properties of a circle to solve geometry problems. Key ideas used are: a radius drawn to a point of contact is perpendicular to the tangent; equal tangents drawn from an external point are of equal length; the power of a point (products of segments formed by secants/tangents) relates external distances to chord/segment lengths; equal chords subtend equal angles and are equidistant from the centre; angle subtended by an arc at the centre is twice the angle subtended on the circumference; and an angle in a semicircle is a right angle. Problem-solving typically reduces to algebra + right-triangle geometry (Pythagoras), similarity of triangles, and simple circle theorems.
Approach to problems: (1) Identify given elements (radii, chords, tangents, external points). (2) Mark right angles (radius to tangent). (3) Use algebraic relations: equal tangent lengths, power-of-a-point relations, chord-distance formula, or angle relations. (4) Solve using Pythagoras or simple algebra. Organise diagrams carefully — many mistakes come from wrong labelling.
- Example 1 — Tangent length and radius: From an external point P, a tangent PT touches a circle at T. The distance from P to the centre O is 13 cm and the radius is 5 cm. Find the length of the tangent PT. Solution: OP = 13, OT = 5 and OT ⟂ PT. In right triangle OPT, PT = √(OP^2 - OT^2) = √(13^2 - 5^2) = √(169 - 25) = √144 = 12 cm.
- Example 2 — Two tangents from an external point: From external point P two tangents touch the circle at A and B. If PA = 8 cm and the distance PB is also tangent from same point, show PB = 8. If the line joining O (centre) and P is 17 cm, find the radius. Solution: Tangents from same external point are equal, so PB = PA = 8. In right triangle OPA: OP^2 = OA^2 + PA^2 ⇒ 17^2 = r^2 + 8^2 ⇒ r^2 = 289 - 64 = 225 ⇒ r = 15 cm.
- Example 3 — Power of a point (secant–tangent): From point P outside a circle a secant cuts the circle at A and B (A nearer P) and a tangent from P touches at T. If PA = 2 cm, PB = 10 cm, find PT. Solution: Power-of-a-point: (PT)^2 = PA × PB. So PT^2 = 2 × 10 = 20 ⇒ PT = √20 = 2√5 cm.
- Example 4 — Chord distance and chord length: In a circle of radius 13 cm, a chord is at a distance 5 cm from the centre. Find the chord length. Solution: If distance from centre to chord = d = 5 and radius r = 13, half-chord length = √(r^2 - d^2) = √(169 - 25) = √144 = 12. So chord length = 2 × 12 = 24 cm.
- Example 5 — Angle in a semicircle: Show that the angle subtended by a diameter AB at any point C on the circle is 90°. (Sketch proof) Solution: Let O be centre. OA and OB are radii, so triangles OCA and OCB are isosceles. Use that ∠AOB = 180° (straight line) and central angle is twice the angle at circumference: ∠ACB = 1/2 ∠AOB = 90°.
- Radius–tangent: radius drawn to point of contact is perpendicular to tangent (OT ⟂ tangent at T).
- Equal tangents: tangents from an external point are equal in length (if PA and PB are tangents from P, then PA = PB).
- Power of a point (tangent–secant): (tangent length)^2 = (external part of secant) × (whole secant) i.e. PT^2 = PA × PB, where A, B are points where secant meets circle (A nearer P).
- Two secants: For two secants PAB and PCD from P, PA × PB = PC × PD.
- Chord length from distance: chord length l = 2 × sqrt(r^2 - d^2), where r is radius and d is distance from centre to chord.
- Angle at centre and circumference: ∠ subtended at centre = 2 × ∠ subtended on circumference by same arc.
Key Concepts
- Circle
- Set of all points in a plane at a fixed distance (radius) from a fixed point (center).
- Radius
- Line segment joining the center of a circle to any point on the circle; all radii are equal.
- Diameter
- A chord passing through the center; its length is twice the radius.
- Chord
- A line segment with both endpoints on the circle.
- Secant
- A line that intersects a circle at two distinct points.
- Tangent
- A line that touches the circle at exactly one point and does not cut it.
- Point of contact (Point of tangency)
- The unique point where a tangent touches the circle.
- Arc
- A continuous part of the circle between two points; can be a minor or major arc.
- Central angle
- Angle with its vertex at the center of the circle subtending an arc.
- Inscribed angle (Angle subtended at the circumference)
- Angle formed by two chords with vertex on the circle; it equals half the central angle subtending the same arc.
- Sector
- Region bounded by two radii and the included arc.
- Segment
- Region between a chord and the corresponding arc (minor or major segment).
- Semicircle
- Arc or region of a circle corresponding to 180° (half the circle).
- Concentric circles
- Two or more circles that share the same center but have different radii.
- Cyclic quadrilateral
- A quadrilateral whose four vertices lie on the same circle; opposite angles are supplementary.
- Power of a point
- For a point P and a circle, product of lengths of segments of any secant through P is constant: if secant meets circle at A,B and at C,D then PA·PB = PC·PD.
- Perpendicular bisector of a chord
- A line through the center of the circle perpendicular to a chord bisects that chord and its subtended arc.
- Equal chords and distance from center
- Equal chords of a circle subtend equal arcs and are equidistant from the center.
- Angles in the same segment
- Angles standing on the same chord and on the same side of it are equal.
- Alternate segment theorem
- The angle between a tangent and a chord through the point of contact equals the angle in the opposite arc.
End-of-Chapter Trial Paper & Test Questions
Topic-wise questions to test your understanding of every concept in this chapter.
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Define a tangent to a circle and state how many tangents can be drawn from a point lying outside the circle. / वृत्त की स्पर्श रेखा को परिभाषित कीजिए और बताइए कि वृत्त के बाहर स्थित किसी बिंदु से कितनी स्पर्श रेखाएँ खींची जा सकती हैं।
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A tangent to a circle is a straight line that touches the circle at exactly one point, called the point of contact; from an external point exactly two tangents can be drawn to a circle. / वृत्त की स्पर्श रेखा वह सीधी रेखा है जो वृत्त को ठीक एक बिंदु पर स्पर्श करती है, जिसे स्पर्श बिंदु कहते हैं; बाहरी बिंदु से वृत्त पर ठीक दो स्पर्श रेखाएँ खींची जा सकती हैं।
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Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact (give the reasoning). / सिद्ध कीजिए कि वृत्त के किसी बिंदु पर खींची गई स्पर्श रेखा स्पर्श बिंदु से होकर जाने वाली त्रिज्या पर लंब होती है (तर्क दीजिए)।
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If the radius OP were not perpendicular to the tangent at P, then the line would meet the circle at a second point, contradicting the fact that a tangent touches the circle at only one point; hence OP must be perpendicular to the tangent. / यदि त्रिज्या OP, P पर स्पर्श रेखा पर लंब न हो, तो रेखा वृत्त को दूसरे बिंदु पर भी मिलेगी, जो इस तथ्य के विरुद्ध है कि स्पर्श रेखा वृत्त को केवल एक बिंदु पर स्पर्श करती है; अतः OP स्पर्श रेखा पर लंब होनी चाहिए।
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From an external point P the distance to the centre O is 13 cm and the radius is 5 cm. Find the length of the tangent PT. / बाहरी बिंदु P से केंद्र O की दूरी 13 सेमी है तथा त्रिज्या 5 सेमी है। स्पर्श रेखा PT की लंबाई ज्ञात कीजिए।
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Since OT ⟂ PT, triangle OPT is right-angled at T, so PT = √(OP² − OT²) = √(13² − 5²) = √(169 − 25) = √144 = 12 cm. / चूँकि OT ⟂ PT, त्रिभुज OPT, T पर समकोण है, अतः PT = √(OP² − OT²) = √(13² − 5²) = √(169 − 25) = √144 = 12 सेमी।
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Why are the lengths of two tangents drawn from an external point to a circle equal? / किसी बाहरी बिंदु से वृत्त पर खींची गई दो स्पर्श रेखाओं की लंबाइयाँ बराबर क्यों होती हैं?
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In the two right triangles formed by the radii to the points of contact, the radii are equal and OP is common, so the triangles are congruent (RHS), making the tangent lengths PA = PB equal. / स्पर्श बिंदुओं तक की त्रिज्याओं से बने दोनों समकोण त्रिभुजों में त्रिज्याएँ बराबर हैं और OP उभयनिष्ठ है, अतः त्रिभुज सर्वांगसम हैं (RHS), जिससे स्पर्श रेखा लंबाइयाँ PA = PB बराबर हो जाती हैं।
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Prove that the perpendicular from the centre of a circle to a chord bisects the chord. / सिद्ध कीजिए कि वृत्त के केंद्र से जीवा पर डाला गया लंब जीवा को समद्विभाजित करता है।
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Let OM ⟂ chord AB. In triangles OAM and OBM, OA = OB (radii), OM is common, and both angles at M are 90°, so the triangles are congruent (RHS); hence AM = MB. / मान लीजिए OM ⟂ जीवा AB। त्रिभुज OAM और OBM में OA = OB (त्रिज्याएँ), OM उभयनिष्ठ है, और M पर दोनों कोण 90° हैं, अतः त्रिभुज सर्वांगसम हैं (RHS); इसलिए AM = MB।
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In a circle of radius 13 cm, a chord is at a distance of 5 cm from the centre. Find the length of the chord. / 13 सेमी त्रिज्या वाले वृत्त में एक जीवा केंद्र से 5 सेमी की दूरी पर है। जीवा की लंबाई ज्ञात कीजिए।
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Half the chord = √(r² − d²) = √(13² − 5²) = √(169 − 25) = √144 = 12 cm, so the chord length = 2 × 12 = 24 cm. / आधी जीवा = √(r² − d²) = √(13² − 5²) = √(169 − 25) = √144 = 12 सेमी, अतः जीवा की लंबाई = 2 × 12 = 24 सेमी।
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Two concentric circles have radii 5 cm and 3 cm. Find the length of a chord of the larger circle that touches the smaller circle. / दो संकेंद्रीय वृत्तों की त्रिज्याएँ 5 सेमी और 3 सेमी हैं। बड़े वृत्त की उस जीवा की लंबाई ज्ञात कीजिए जो छोटे वृत्त को स्पर्श करती है।
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The chord is tangent to the smaller circle, so the radius (3 cm) is perpendicular to it at the midpoint; half-chord = √(5² − 3²) = √16 = 4 cm, hence chord = 8 cm. / जीवा छोटे वृत्त की स्पर्श रेखा है, अतः त्रिज्या (3 सेमी) मध्य बिंदु पर उस पर लंब है; आधी जीवा = √(5² − 3²) = √16 = 4 सेमी, अतः जीवा = 8 सेमी।
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A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. / एक चतुर्भुज ABCD एक वृत्त के परिगत खींचा गया है। सिद्ध कीजिए कि AB + CD = AD + BC।
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Using equal tangent lengths from each vertex, adding the tangent segments gives AB + CD = (AP + PB) + (CR + RD) and AD + BC = (AS + SD) + (BQ + QC); since tangents from each vertex are equal, both sums equal, so AB + CD = AD + BC. / प्रत्येक शीर्ष से बराबर स्पर्श रेखा लंबाइयों का प्रयोग करते हुए, स्पर्श खंडों को जोड़ने पर AB + CD = (AP + PB) + (CR + RD) तथा AD + BC = (AS + SD) + (BQ + QC); चूँकि प्रत्येक शीर्ष से स्पर्श रेखाएँ बराबर हैं, दोनों योग बराबर हैं, अतः AB + CD = AD + BC।
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