Chapter 16 — Tangents and Secants to a Circle

Take a circle with centre O and radius r drawn on a sheet of paper, and lay a straight edge across the sheet in different positions. Only three things can happen. The line may pass entirely outside the circle and have no point in common with it; we call this a non-intersecting line. The line may cut through the circle so that it has two points in common with it; such a line is called a secant, and the part of it that lies inside the circle, joining the two common points, is a chord. Finally the line may just graze the circle so that it has exactly one point in common; this line is a tangent, and the single common point is called the point of contact or point of tangency.

There is a neat way to say which case we are in. Drop a perpendicular from the centre O to the line and let its length be d. If d is greater than r the line misses the circle, because every point of the line is at distance at least d from O and so lies outside. If d is less than r the perpendicular foot lies inside the circle, so the line must cross the circle on both sides of the foot: the line is a secant. If d equals r exactly, the foot of the perpendicular is itself on the circle and no other point of the line can be, since every other point is farther from O than d. That is the tangent case. So the three positions correspond to d > r, d < r and d = r.

A picture from daily life helps. A bicycle wheel standing on a road touches the road at exactly one point; the road is a tangent to the wheel. A ruler laid across a coin so that it crosses the rim twice is a secant. A ruler held above the coin without touching it is non-intersecting. The word tangent comes from the Latin tangere, to touch, and the word secant from secare, to cut.

Two more facts are worth noting now. A tangent can be drawn at every point of a circle, and there is only one tangent at each point, because the tangent must be the unique line through that point perpendicular to the radius (this is the first theorem of the chapter). Also, a secant that is moved gradually away from the centre has its two intersection points come closer and closer together until they coincide; at that instant the secant has become the tangent. This idea is taken up in the next topic.

Draw a circle with centre O and a secant through two points A and B of the circle. Now keep the point A fixed and let the point B move along the circle towards A. As B approaches A the secant AB turns about A, and the chord AB becomes shorter and shorter. When B finally reaches A the two points of intersection coincide, the chord has shrunk to a single point, and the line no longer cuts the circle: it merely touches it at A. This final line is the tangent at A, and it is described as the limiting position of the secant AB when B tends to A.

The same thing can be seen by moving the secant parallel to itself. Start with a secant close to the centre and slide it outward keeping its direction fixed. Its two intersection points move towards each other along the circle, and at the moment the distance of the line from the centre becomes equal to the radius the two points merge into one. Slide the line any further and it leaves the circle altogether. So the tangent is the boundary case between the secant and the non-intersecting line.

This viewpoint gives a useful description: a tangent to a circle is a line that meets the circle in exactly one point, and the point of contact can be thought of as two coincident points of intersection. This is why algebra, when it is used to find where a line meets a circle, gives a repeated root in the tangent case and two distinct roots in the secant case. The student will use that idea in the Intermediate course; here it is enough to understand the picture.

An activity makes the idea concrete. Fix a wire through a hole in a circular cardboard disc at a point A of its rim, and rotate the wire about A. In every position but one the wire crosses the rim at a second point; in exactly one position it crosses only at A. That single position is the tangent. Also notice that in that position the wire is perpendicular to the radius OA, which is the theorem proved in the next topic.

Two properties follow directly from this picture. First, at each point of the circle there is one and only one tangent. Second, every tangent has exactly one point of contact, so a line that meets a circle in two points, however close together, is still a secant and not a tangent. The student should use the words correctly: a line touches a circle (tangent) or cuts a circle (secant); it does not do both.

Theorem. The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given: a circle with centre O and a tangent XY touching the circle at the point P. To prove: OP ⊥ XY.

Proof. Take any point Q on the line XY other than P and join OQ. Since XY is a tangent, it meets the circle only at P; so Q is not on the circle. Could Q be inside the circle? If it were, then the line XY, passing through a point inside the circle, would have to cut the circle at two points and would be a secant, contradicting the fact that it is a tangent. Hence Q lies outside the circle, and therefore OQ > r, that is, OQ > OP. This is true for every point Q of XY except P. So OP is the shortest of all the segments joining O to points of the line XY. But the shortest segment from a point to a line is the perpendicular from that point to the line. Hence OP ⊥ XY. This proves the theorem.

Two remarks make the theorem more useful. First, the line through the centre perpendicular to the tangent passes through the point of contact, and the perpendicular to the tangent drawn at the point of contact passes through the centre. So whenever a tangent appears in a figure, the student should immediately draw the radius to the point of contact and mark the right angle; almost every tangent problem depends on that right angle. Second, the converse is also true: a line drawn through a point of the circle perpendicular to the radius at that point is a tangent, because every other point of that line is farther from the centre than the foot of the perpendicular, so it lies outside the circle.

The theorem is the reason a wheel resting on the ground stands with its axle exactly above the point of contact, and the reason a stone released from a whirling sling flies off along the tangent, which is at right angles to the string. It also gives a quick way to compute lengths: if a tangent from a point P touches the circle at T, then triangle OTP is right-angled at T, so by Pythagoras PT² = OP² − OT² = OP² − r².

Worked example. A tangent PT is drawn to a circle of radius 5 cm from a point P at distance 13 cm from the centre O. Since ∠OTP = 90°, PT² = 13² − 5² = 169 − 25 = 144, so PT = 12 cm. Conversely, if the tangent length is 8 cm and the radius is 6 cm, then OP = √(64 + 36) = 10 cm.

Because the tangent at a point is the line through that point perpendicular to the radius, drawing a tangent at a given point of a circle reduces to drawing a perpendicular, which the student already knows how to do.

Construction: to draw the tangent to a circle at a given point P on it, when the centre O is known. Step 1: join O to P. Step 2: at P draw a line XY perpendicular to OP, by the usual compass construction (with P as centre draw arcs cutting the line OP produced on either side of P, then from those two points draw equal arcs meeting at a point, and join that point to P). Step 3: the line XY is the required tangent. Justification: XY passes through the point P of the circle and is perpendicular to the radius OP, so by the converse of the tangent theorem it is the tangent at P.

Construction when the centre is not given. Sometimes only the circle and the point P are drawn. Then draw any chord PQ from P, and take a point R on the major arc PQ. Join PR and QR. At P construct the angle ∠QPX equal to ∠PRQ, on the opposite side of PQ from R. The line PX is the tangent at P. The justification uses the alternate segment property, which the student meets later; for the board examination the construction with a known centre is the important one. An easier alternative when the centre is not given is to find the centre first, by drawing the perpendicular bisectors of any two chords, and then use the first construction.

The student should be able to draw the tangents at the ends of a diameter as well. The tangents at the two ends A and B of a diameter AB are both perpendicular to AB, and two lines perpendicular to the same line are parallel. So tangents at the ends of a diameter are parallel. Conversely, if two tangents to a circle are parallel, the points of contact are the ends of a diameter, because the radii to the points of contact are both perpendicular to the same direction and therefore lie along one line through the centre.

A related fact concerns the angle between a chord and the tangent at one end of it. If PQ is a chord and the tangent at P makes angle θ with PQ, then the angle subtended by PQ at the centre is 2θ, because triangle OPQ is isosceles with the angles at P and Q each equal to 90° − θ, leaving 180° − 2(90° − θ) = 2θ at O. This gives a way of finding central angles from tangent angles in examination problems.

Fix a circle with centre O and choose a point P. The number of tangents that can be drawn to the circle passing through P depends only on where P lies.

Case 1: P inside the circle. Every line through an interior point must cross the circle at two points, because it passes from inside to outside on both sides of P. So every line through P is a secant, and no tangent can be drawn from an interior point. The centre is a special interior point: every line through it is a diameter.

Case 2: P on the circle. There is exactly one tangent through P, namely the line through P perpendicular to OP. Any other line through P makes an angle other than 90° with OP, so it comes closer to the centre than r on one side of P and must cut the circle again.

Case 3: P outside the circle. Exactly two tangents can be drawn from P. To see why, consider the lines through P. Those that pass close to the centre are secants; those that pass far from it miss the circle. Rotating a line about P from a secant position to a non-intersecting position, it must pass through a tangent position exactly once on each side of the line PO. So there are two tangents. The two points of contact are the ends of a chord, called the chord of contact of P, which is perpendicular to PO and bisected by it.

The segment from P to the point of contact of a tangent is called the length of the tangent from P. From an external point there are therefore two such lengths, and the next theorem says they are equal. Note the distinction in language: a tangent is a line and has no length; the length of the tangent is the distance from the external point to the point of contact.

This counting also tells us something about the number of common tangents of two circles. Two circles that intersect at two points have two common tangents; two circles touching externally have three; two circles lying entirely outside each other have four (two direct and two transverse); two circles touching internally have one; and one circle lying inside another without touching has none. These facts are asked as short questions in the board examination and can be reasoned out from a figure.

Worked example. A point P is at a distance of 10 cm from the centre of a circle of radius 6 cm. Since 10 > 6 the point is outside, so two tangents can be drawn, and each has length √(100 − 36) = 8 cm.

Theorem. The lengths of tangents drawn from an external point to a circle are equal.

Given: a circle with centre O, an external point P, and tangents PA and PB touching the circle at A and B. To prove: PA = PB.

Proof. Join OA, OB and OP. By the previous theorem the tangent is perpendicular to the radius at the point of contact, so ∠OAP = 90° and ∠OBP = 90°. Now compare the right triangles OAP and OBP. OA = OB, being radii of the same circle. OP is common to both. So the hypotenuse and one side of one right triangle are equal to the hypotenuse and one side of the other, and by the RHS congruence rule, triangle OAP ≅ triangle OBP. Corresponding parts of congruent triangles are equal, so PA = PB. This proves the theorem.

The congruence gives more than the equality of lengths, and the extra facts are used constantly in problems. (i) ∠OPA = ∠OPB: the line joining the external point to the centre bisects the angle between the two tangents. (ii) ∠AOP = ∠BOP: the line PO also bisects the angle AOB between the two radii. (iii) Since PA = PB, triangle PAB is isosceles, so PO, being the bisector of its vertical angle, is the perpendicular bisector of the chord of contact AB. (iv) In the quadrilateral OAPB the angles at A and B are 90°, so ∠AOB + ∠APB = 180°: the angle between the two tangents is supplementary to the angle subtended by the chord of contact at the centre.

Worked example 1. Two tangents PA and PB are drawn from a point P to a circle with centre O, and ∠APB = 60°. Then ∠AOB = 180° − 60° = 120°, and each of ∠OPA and ∠OPB is 30°. If the radius is 6 cm, then in the right triangle OAP, tan 30° = OA/PA, so PA = 6/tan 30° = 6√3 cm, and OP = OA/sin 30° = 12 cm.

Worked example 2. From a point 17 cm from the centre of a circle of radius 8 cm, the two tangents have equal length √(289 − 64) = √225 = 15 cm each.

A frequent examination question asks the student to prove the theorem, so the steps above should be reproduced with the figure, the given, the statement to be proved and the reasons written beside each step.

Construction: to draw the two tangents to a circle from a point P outside it.

Step 1: draw the circle with centre O and mark the external point P. Join OP. Step 2: bisect the segment OP and let M be its midpoint (draw the perpendicular bisector of OP with compass arcs from O and P). Step 3: with M as centre and MO as radius draw a circle; it passes through O and P. Step 4: this circle cuts the given circle at two points A and B. Step 5: join PA and PB. These are the required tangents.

Justification. Join OA. The segment OP is a diameter of the circle drawn in Step 3, and A is a point on that circle, so ∠OAP is an angle in a semicircle and equals 90°. Hence PA is perpendicular to the radius OA at the point A of the given circle, and therefore PA is a tangent by the converse of the tangent theorem. The same argument with B shows that PB is a tangent. Because the two circles meet in exactly two points, exactly two tangents are obtained, in agreement with the previous topic.

After the construction the student can measure PA and PB and check that they are equal, confirming the theorem on equal tangents, and can compute the expected length from √(OP² − r²) to check the drawing.

Worked construction. Draw a circle of radius 3 cm and a point P at 7 cm from its centre; construct the tangents and measure their lengths. The expected length is √(49 − 9) = √40 ≈ 6.3 cm. In practice a careful drawing gives 6.3 cm.

A variation asked in the examination: draw a pair of tangents to a circle of radius 4 cm which are inclined to each other at 60°. Since ∠APB = 60°, the angle AOB at the centre is 120°. So draw the circle, draw any radius OA, construct ∠AOB = 120°, and at A and B draw the perpendiculars to OA and OB; they meet at P and are the required tangents. Alternatively use OP = r/sin 30° = 8 cm: mark P at 8 cm from O and apply the standard construction.

The examination also asks for the tangents to a circle from a point when the centre is not marked. Then first locate the centre as the intersection of the perpendicular bisectors of two chords, and proceed as above. Every construction answer must carry the steps of construction in words and a short justification; the marks are divided between the drawing and the writing.

The two theorems yield a family of standard results, each of which has been asked in the board examination. The method is always the same: draw the radius to each point of contact, mark the right angles, and use the equality of tangents from each external point.

Result 1: a quadrilateral circumscribing a circle. If a quadrilateral ABCD is drawn so that all four sides touch a circle at P, Q, R, S (on AB, BC, CD, DA respectively), then AB + CD = BC + DA. Proof. Tangents from A: AP = AS. From B: BP = BQ. From C: CR = CQ. From D: DR = DS. Adding, AP + BP + CR + DR = AS + BQ + CQ + DS, that is, AB + CD = BC + DA. A parallelogram circumscribing a circle is therefore a rhombus, since opposite sides are equal and their sums are equal, forcing all sides equal.

Result 2: the angle subtended by opposite sides at the centre. If ABCD circumscribes a circle with centre O, then ∠AOB + ∠COD = 180°. This follows because OP, OQ, OR, OS bisect the angles between the tangent pairs, and the eight small angles at O pair up.

Result 3: a tangent at a point on a chord of contact. If PA and PB are tangents from P and a third tangent at a point C on the minor arc AB meets PA at D and PB at E, then the perimeter of triangle PDE equals 2PA, because DC = DA and EC = EB, so PD + DE + EP = PD + DA + EB + EP = PA + PB = 2PA.

Result 4: the incircle of a right triangle. If a circle of radius r is inscribed in a right triangle with legs a, b and hypotenuse c, then r = (a + b − c)/2, because the two tangents from the right-angled vertex together with the two radii form a square of side r, and the remaining tangent lengths on the legs add to the hypotenuse.

Result 5: parallel tangents. Two parallel tangents are at distance 2r apart and their points of contact are the ends of a diameter. If a third tangent cuts them at A and B, then ∠AOB = 90°, because OA and OB bisect the two angles that sum to 180° at the two ends of the transversal.

Worked example. A triangle with sides 8 cm, 15 cm and 17 cm (a right triangle, since 64 + 225 = 289) has an inscribed circle; its radius is (8 + 15 − 17)/2 = 3 cm. Another: the sides of a quadrilateral circumscribing a circle are AB = 6 cm, BC = 7 cm, CD = 4 cm; then AD = AB + CD − BC = 6 + 4 − 7 = 3 cm.

A secant cuts a circle at two points A and B, and the part AB inside the circle is a chord. The chord divides the circular region into two parts. Each part is a segment of the circle: the region bounded by the chord and the arc on that side. The smaller region, lying on the side of the minor arc, is called the minor segment; the larger region, lying on the side of the major arc, is the major segment. If the chord is a diameter the two segments are equal and each is a semicircle.

Along with the segment we need the sector. Two radii OA and OB together with the arc AB bound a region called a sector; the angle ∠AOB between the radii is the angle of the sector or central angle. The sector on the side of the minor arc is the minor sector and the other one is the major sector. The difference between a sector and a segment is that a sector has the centre as a vertex and is bounded by two radii, while a segment is bounded by a chord and contains no radius.

The relation between the two figures is the key to the whole calculation of areas: the minor sector OAB is made up of the triangle OAB and the minor segment on chord AB. Hence

Area of minor segment = Area of minor sector OAB − Area of triangle OAB.

And since the whole circle is divided by the chord into the two segments,

Area of major segment = Area of circle − Area of minor segment.

The student must be able to identify these regions in a figure and to state which quantities are needed to compute them: the radius r and the central angle x subtended by the chord. The central angle is often found from the geometry of the figure: a chord equal to the radius subtends 60° (equilateral triangle), a chord equal to r√2 subtends 90° (isosceles right triangle), and a chord equal to r√3 subtends 120°. Conversely, if the angle is given, the chord is 2r sin(x/2), which the student can also find by dropping the perpendicular from O to the chord and using trigonometry.

A concrete example: in a circle of radius 10 cm a chord subtends 90° at the centre. The chord is 10√2 cm long, the minor sector is a quarter of the circle, the triangle OAB is a right isosceles triangle of legs 10 cm, and the minor segment is the quarter circle with that triangle removed. The numbers are worked out in the topic on segment area. The shape of a segment is familiar: it is the shape of a slice cut from a round bread with one straight cut, or the shape of the water surface in a partly filled horizontal pipe.

The area of a whole circle of radius r is πr² and the central angle of the whole circle is 360°. A sector with central angle x° is a fraction x/360 of the circle, because the area of a sector is proportional to its angle: doubling the angle doubles the area. So

Area of a sector of angle x° = (x/360) × πr².

Similarly the circumference 2πr corresponds to 360°, so the arc of a sector of angle x° has

Length of arc = (x/360) × 2πr.

Combining the two, area of sector = ½ × (length of arc) × r, since (x/360)πr² = ½ × (x/360)2πr × r. This form is useful when the arc length is given instead of the angle. The perimeter of a sector is the arc length plus the two radii: (x/360)2πr + 2r.

Worked example 1. Find the area of a sector of angle 60° in a circle of radius 6 cm (π = 3.14). Area = (60/360) × 3.14 × 36 = (1/6) × 113.04 = 18.84 cm². Arc length = (60/360) × 2 × 3.14 × 6 = (1/6) × 37.68 = 6.28 cm. Perimeter of the sector = 6.28 + 12 = 18.28 cm.

Worked example 2. The minute hand of a clock is 14 cm long. Find the area swept in 5 minutes (π = 22/7). In 60 minutes the hand turns 360°, so in 5 minutes it turns 30°. Area swept = (30/360) × (22/7) × 14 × 14 = (1/12) × 616 = 51.33 cm².

Worked example 3. A sector has area 77 cm² in a circle of radius 14 cm; find its angle. (x/360) × (22/7) × 196 = 77, so (x/360) × 616 = 77, giving x/360 = 1/8 and x = 45°.

Worked example 4. An umbrella has 8 ribs equally spaced, each 45 cm long, and is a flat circle when opened. The area of cloth between two consecutive ribs is a sector of angle 360/8 = 45°: (45/360) × (22/7) × 45 × 45 = (1/8) × 6364.29 = 795.54 cm².

A few points about method. Always write the fraction x/360 first and cancel before multiplying; use π = 22/7 when the radius is a multiple of 7 and 3.14 otherwise, unless the question says which to use. For the major sector, use the angle 360° − x, or subtract the minor sector from the circle. When the angle is in radians the fraction becomes x/2π, but this chapter works in degrees.

To find the area of the minor segment cut off by a chord AB that subtends an angle x° at the centre of a circle of radius r, we subtract the triangle OAB from the sector OAB.

The area of the triangle OAB can be found in two ways. If x is 60°, 90° or 120° the triangle is equilateral, right isosceles or has a 30°-30°-120° shape, and its area is (√3/4)r², ½r², or (√3/4)r² respectively. In general, drop the perpendicular OM from O to AB; M is the midpoint of AB, ∠AOM = x/2, so AM = r sin(x/2) and OM = r cos(x/2), and the area is ½ × AB × OM = ½ × 2r sin(x/2) × r cos(x/2) = r² sin(x/2) cos(x/2), which equals ½r² sin x.

Area of minor segment = (x/360)πr² − ½ r² sin x. The major segment is πr² minus this.

Worked example 1. A chord subtends 90° at the centre of a circle of radius 10 cm (π = 3.14). Sector = (90/360) × 3.14 × 100 = 78.5 cm². Triangle OAB is right-angled at O with legs 10 cm: area = ½ × 10 × 10 = 50 cm². Minor segment = 78.5 − 50 = 28.5 cm². Major segment = 314 − 28.5 = 285.5 cm².

Worked example 2. A chord subtends 120° at the centre of a circle of radius 12 cm (π = 3.14, √3 = 1.732). Sector = (120/360) × 3.14 × 144 = 150.72 cm². In triangle OAB, OM = 12 cos 60° = 6 cm and AM = 12 sin 60° = 6√3 cm, so AB = 12√3 cm and area = ½ × 12√3 × 6 = 36√3 = 62.35 cm². Minor segment = 150.72 − 62.35 = 88.37 cm².

Worked example 3. A chord of length equal to the radius, in a circle of radius 6 cm, cuts off a segment. The angle is 60°, sector = (60/360) × 3.14 × 36 = 18.84 cm², triangle is equilateral of side 6: (√3/4) × 36 = 9 × 1.732 = 15.59 cm², segment = 3.25 cm².

Worked example 4. A chord 10 cm long is at distance 5√3 cm from the centre of a circle of radius 10 cm. Then cos(x/2) = 5√3/10 = √3/2, so x/2 = 30° and x = 60°; the segment is (60/360)(3.14)(100) − (√3/4)(100) = 52.33 − 43.30 = 9.03 cm².

Points of method: state the sector area, the triangle area and the subtraction separately, each with units; keep √3 and π as symbols until the last step to avoid rounding error; and check that the minor segment is smaller than the sector.

The board examination combines the results of this chapter into problems on the area of shaded regions. The standard ones are these.

(a) A square with four quarter circles at its corners. A square of side 14 cm has quarter circles of radius 7 cm drawn at each corner. The four quarter circles make one full circle of area (22/7) × 49 = 154 cm², and the shaded part between them is 196 − 154 = 42 cm².

(b) A circle inscribed in a square. A circle of radius 7 cm is drawn in a square of side 14 cm; the four corner regions together have area 196 − 154 = 42 cm².

(c) A square inscribed in a circle. A square of side 10 cm is drawn inside a circle; the diagonal 10√2 cm is the diameter, so r = 5√2 cm and the area between them is 3.14 × 50 − 100 = 57 cm².

(d) A segment shaded in a circle. This is the sector-minus-triangle computation of the previous topic.

(e) A semicircle drawn on a side of a right triangle. The shaded region between the semicircle and the triangle is found by subtracting the triangle from the semicircle or by adding and subtracting appropriate pieces.

Summary of the chapter. A line is non-intersecting, a secant or a tangent according as its distance from the centre is greater than, less than or equal to the radius. The tangent at a point is perpendicular to the radius through that point, and the line through a point of the circle perpendicular to the radius is the tangent. From an internal point no tangent, from a point on the circle one, and from an external point exactly two tangents can be drawn. The two tangents from an external point are equal, the line to the centre bisects the angle between them, and the angle between the tangents is supplementary to the central angle. A quadrilateral circumscribing a circle has AB + CD = BC + DA. Tangents are constructed at a point using the perpendicular, and from an external point using the circle on OP as diameter. A chord divides the circle into a minor and a major segment; area of sector = (x/360)πr², length of arc = (x/360)2πr, and area of minor segment = sector − triangle = (x/360)πr² − ½r² sin x.

Question patterns. One-mark questions ask how many tangents can be drawn from a given point, the number of common tangents of two circles, or the angle between a tangent and a radius. Two- and four-mark questions ask for the proof of one of the two theorems, the length of a tangent from a point at a given distance, or the fourth side of a circumscribed quadrilateral. Construction questions ask for the tangents from an external point, or two tangents inclined at a given angle, with steps and justification. Area questions ask for the area of a segment or of a shaded region with π given as 22/7 or 3.14. Marks are lost most often by forgetting to draw the radius to the point of contact, by using the wrong angle for the sector, and by dropping units; a neat labelled figure prevents all three.