What is a tangent to a circle?

A tangent is a line that meets a circle at exactly one point, called the point of contact. It is one of only three things a line can do with respect to a circle: miss it completely (a non-intersecting line, no common point), cut it (a secant, two common points) or touch it (a tangent, exactly one common point). A tangent is the limiting case of a secant, reached when the two ends of the corresponding chord come together.

How many tangents can be drawn to a circle from a point?

It depends on where the point is. From a point INSIDE the circle, no tangent can be drawn at all, because every line through an interior point cuts the circle twice. From a point ON the circle there is exactly one tangent, the line perpendicular to the radius at that point. From a point OUTSIDE the circle there are exactly two tangents, and by Theorem 10.2 their lengths from that point are equal.

What are the two theorems in Class 10 Maths Chapter 10?

Theorem 10.1 says the tangent at any point of a circle is perpendicular to the radius through the point of contact. Theorem 10.2 says the lengths of the two tangents drawn from an external point to a circle are equal. Every question in Exercise 10.1 and Exercise 10.2 is solved using one or both of these, usually alongside the angle sum of a triangle or quadrilateral, the Pythagoras Theorem, or a congruence rule.

Why is the angle between two tangents supplementary to the angle at the centre?

Join the centre O to the external point P and to the two points of contact A and B. In the quadrilateral OAPB the angles at A and at B are both right angles, because each is the angle between a tangent and the radius drawn to its point of contact (Theorem 10.1). Those two right angles use up 180 degrees of the quadrilateral's 360-degree angle sum, so the remaining two angles, the angle APB between the tangents and the angle AOB at the centre, must add up to 180 degrees.

How many exercises does Class 10 Maths Chapter 10 Circles have?

Two. Exercise 10.1 has 4 questions and comes at the end of Section 10.2, before the chapter's first worked Example; Exercise 10.2 has 13 questions and comes at the end of Section 10.3, after Examples 1 to 3. In the rationalised 2026-27 book, Exercise 10.2 Q13 is followed directly by the chapter Summary.

NCERT Solutions Class 10 Maths Ch 10 Circles and Tangents

3 Examples + 17 Questions Solved

Chapter 10: Circles — NCERT Solutions (rationalised syllabus 2026–27), with all 3 worked Examples from the chapter plus every question in Exercise 10.1 and Exercise 10.2. Class 9 covered what a circle is and the terms chord, arc, segment and sector. This chapter asks a narrower question: what can a straight line do with respect to a circle? There are only three possibilities, shown below.

Fig. 10.1 — a line PQ and a circle: (i) no common point (non-intersecting line), (ii) two common points A and B (a secant), (iii) exactly one common point A (a tangent)

If the line and the circle have no common point, the line is called non-intersecting. If they have two common points, the line is a secant, and the part of it inside the circle is a chord. If they have exactly one common point, the line is a tangent and that point is the point of contact — the tangent is said to touch the circle there. A tangent is really the limiting position of a secant: slide a secant outwards and its two intersection points move closer together until they coincide, and the secant has become a tangent.

The whole chapter then rests on just two theorems, both stated in the box below. Every single answer on this page uses one of them, usually together with something already familiar — the angle sum of a triangle or a quadrilateral, the Pythagoras Theorem, or one of the congruence rules. Note the shape of the chapter: NCERT prints Exercise 10.1 before the first worked Example exists, because Section 10.2 contains only Theorem 10.1 and two activities, so that exercise comes first below, exactly as in the book. Exercise 10.2 Q13 is followed directly by the chapter Summary, with no further section or question after it.

Key Theorems Used on This Page

Theorem 10.1 — the tangent is perpendicular to the radius: The tangent at any point of a circle is perpendicular to the radius through the point of contact. If the tangent XY touches the circle of centre O at P, then
\(OP \perp XY\)
This is the single most-used fact on this page. Exercise 10.2 Q5 proves the statement in the other direction: the perpendicular raised at the point of contact has to pass through the centre — which is what lets you construct a tangent.
Remark on Theorem 10.1: At any point of a circle there is one and only one tangent. The line containing the radius through the point of contact is sometimes called the ‘normal’ to the circle at that point.
Theorem 10.2 — equal tangent lengths from an external point: The lengths of the two tangents drawn from an external point to a circle are equal. If PQ and PR are the tangents from P, touching at Q and R, then
\(PQ = PR\)
The proof compares the right triangles OQP and ORP: OQ = OR are radii, OP is common, and both have a right angle at the point of contact, so they are congruent by RHS and PQ = PR by CPCT.
Remark on Theorem 10.2: The same congruence also gives \(\angle OPQ = \angle OPR\), so OP bisects the angle between the two tangents — that is, the centre lies on the bisector of the angle QPR. Exercise 10.2 Q3 uses exactly this.
The length of the tangent: The length of the tangent from an external point P to a circle of centre O and radius \(r\) is the distance from P to the point of contact. Since the triangle is right-angled there,
\(\text{length of tangent} = \sqrt{OP^{2}-r^{2}}\)
This one line answers Exercise 10.1 Q3 and Exercise 10.2 Q1 and Q6.

NCERT proves only these two theorems in this chapter, and the Summary on page 153 lists exactly three points: the meaning of a tangent, Theorem 10.1 and Theorem 10.2. Everything else on this page is built from them together with results already known from earlier classes — the angle sum of a triangle and of a quadrilateral, the Pythagoras Theorem, the congruence rules, and the fact that the perpendicular from the centre of a circle to a chord bisects that chord.

Exercise 10.1

How many tangents can a circle have?

Answer

A tangent is a line that meets the circle at exactly one point, and that point is called the point of contact. So counting the tangents of a circle is the same as counting the possible points of contact.

Step 1 — every point of the circle has a tangent. Take any point P on the circle and join the centre O to it. The line through P perpendicular to OP meets the circle only at P, so it is a tangent at P. Therefore no point of the circle is left without a tangent.

Step 2 — each point has only one. By Theorem 10.1 every tangent at P is perpendicular to OP, and through the point P there is only one line perpendicular to OP. So the tangent at P is unique.

Step 3 — different points give different tangents. If two different points of contact gave the same line, that line would meet the circle at two points and would be a secant, not a tangent. So distinct points of contact always give distinct tangents.

Step 4 — count them. A circle contains infinitely many points, and by Steps 1–3 each of them contributes exactly one tangent, all different from one another.

Hence a circle can have infinitely many tangents — one at each point of the circle.

Fill in the blanks:(i) A tangent to a circle intersects it in ________ point(s).(ii) A line intersecting a circle in two points is called a ________.(iii) A circle can have ________ parallel tangents at the most.(iv) The common point of a tangent to a circle and the circle is called ________.

Answer

(i) one. Section 10.2 defines a tangent as a line that intersects the circle at only one point — that is what separates it from the other two cases of Fig. 10.1: a non-intersecting line has no common point with the circle, and a secant has two.

(ii) secant. This is the middle case of Fig. 10.1: the line PQ meets the circle at the two points A and B, and the part of it inside the circle is the chord AB. A tangent is the limiting position of a secant, reached when the two ends of that chord come together.

(iii) two. Fix any direction. A tangent in that direction has to sit at a perpendicular distance of exactly one radius from the centre, and there are only two positions at that distance — one on each side of the centre. Those two are the tangents at the two ends of the diameter perpendicular to the chosen direction, and Exercise 10.2 Q4 below proves they really are parallel. So the largest number of mutually parallel tangents a circle can have is two.

(iv) point of contact. Section 10.2 names the common point of the tangent and the circle the point of contact, and says the tangent touches the circle there.

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:(A) 12 cm(B) 13 cm(C) 8.5 cm(D) \(\sqrt{119}\) cm

Exercise 10.1 Q3 — tangent PQ touches the circle at P; OP = 5 cm is the radius and OQ = 12 cm, so ∠OPQ = 90°

Answer

Step 1 — find the right angle. PQ is a tangent to the circle at P and OP is the radius drawn to the point of contact. By Theorem 10.1 the tangent is perpendicular to that radius, so

\(\angle OPQ = 90^{\circ}\)

Therefore △OPQ is a right triangle, right-angled at P, with OQ as its hypotenuse.

Step 2 — apply the Pythagoras Theorem. In △OPQ,

\(OQ^{2} = OP^{2} + PQ^{2}\)

\(12^{2} = 5^{2} + PQ^{2}\)

\(144 = 25 + PQ^{2}\)

Step 3 — solve for PQ.

\(PQ^{2} = 144 - 25 = 119\)

\(PQ = \sqrt{119}\text{ cm}\)

So the correct option is (D) \(\sqrt{119}\) cm.

A quick sanity check on the wrong options. PQ is a leg of the right triangle and OQ = 12 cm is the hypotenuse, so PQ must be less than 12 cm. That rules out (A) 12 cm and (B) 13 cm immediately. And \(\sqrt{119}\approx 10{\cdot}91\) cm, which is not 8.5 cm, so (C) is out too.

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Exercise 10.1 Q4 — the required construction: the tangent touches the circle at exactly one point T, the secant cuts it at two points, and both are parallel to the given line

Answer

The answer to this question is the construction itself, shown above. Here is how each line is drawn, and why it does what the question asks.

Step 1 — the given line. Draw any line and call it the given line. Every other line in this construction has to be parallel to it.

Step 2 — the circle. Mark a point O away from the given line and draw a circle with centre O and any convenient radius, taking care that the circle does not cut the given line (that keeps the picture uncluttered; it is not essential).

Step 3 — the tangent. Drop the perpendicular from O to the given line and let it meet the circle at the point T nearer the line. Now draw the line through T parallel to the given line. That line is perpendicular to OT at T, and OT is a radius — so by the result of Exercise 10.2 Q5 below (a line perpendicular to a radius at its outer end is the tangent there) it touches the circle at T and nowhere else. It is the required tangent, and it is parallel to the given line by construction.

Step 4 — the secant. Draw a second line parallel to the given line, but this time at a perpendicular distance from O that is less than the radius. Because a chord exists at every distance from the centre smaller than the radius, this line cuts the circle at two points, so it is a secant.

The idea behind Steps 3 and 4. For a fixed direction, whether a line touches, cuts or misses the circle depends only on its perpendicular distance \(d\) from the centre: \(d > r\) gives a non-intersecting line, \(d = r\) gives the tangent, and \(d < r\) gives a secant. Sliding a line parallel to itself changes \(d\) and nothing else, which is exactly why one direction can supply all three cases.

Worked Examples (Examples 1–3)

Example 1

Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

Fig. 10.8 — concentric circles C₁ and C₂ with centre O; the chord AB of the larger circle touches the smaller circle at P

Solution

What is given. Two concentric circles — the larger C₁ and the smaller C₂ — with the same centre O, and a chord AB of C₁ which touches C₂ at the point P (Fig. 10.8).

What is to be proved. AP = BP, that is, P is the midpoint of the chord AB.

Step 1 — get a right angle at P. Join OP. The line AB touches the smaller circle C₂ at P, so AB is a tangent to C₂ and OP is the radius of C₂ drawn to the point of contact. By Theorem 10.1,

\(OP \perp AB\)

Step 2 — now look at the larger circle. In the circle C₁, the segment AB is a chord and OP is the perpendicular dropped from the centre O onto it. The perpendicular from the centre of a circle to a chord bisects the chord, so AP = BP and we are done. Here is that standard result proved in full, so nothing is taken on trust.

Step 3 — the proof of the bisecting property. Join OA and OB, and compare the two triangles OPA and OPB:

\(OA = OB\ \text{(radii of the larger circle }C_1\text{)}\)

\(OP = OP\ \text{(common side)}\)

\(\angle OPA = \angle OPB = 90^{\circ}\ \text{(from Step 1)}\)

These are right triangles (right-angled at P by Step 1) in which the hypotenuses OA and OB are equal and one pair of sides OP is common, so by the RHS congruence rule

\(\triangle OPA \cong \triangle OPB\ \text{(RHS)}\)

Corresponding parts of congruent triangles are equal, so

\(AP = BP\ \text{(CPCT)}\)

Hence the chord AB of the larger circle is bisected at its point of contact P with the smaller circle.

Example 2

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that \(\angle PTQ = 2\,\angle OPQ\).

Fig. 10.9 — tangents TP and TQ drawn to a circle with centre O from the external point T; P and Q are the points of contact

Solution

What is given. A circle with centre O, an external point T, and the two tangents TP and TQ from T, where P and Q are the points of contact (Fig. 10.9).

Step 1 — name the angle we are chasing. Let

\(\angle PTQ = \theta\)

Step 2 — the triangle TPQ is isosceles. By Theorem 10.2 the two tangent lengths from an external point are equal, so TP = TQ. In a triangle, angles opposite equal sides are equal, so the two base angles of △TPQ are equal:

\(\angle TPQ = \angle TQP\)

Step 3 — find those base angles. The three angles of △TPQ add up to \(180^{\circ}\), so the two equal base angles share whatever is left after \(\theta\):

\(\angle TPQ + \angle TQP + \angle PTQ = 180^{\circ}\)

\(\angle TPQ = \angle TQP = \tfrac{1}{2}\left(180^{\circ}-\theta\right) = 90^{\circ}-\tfrac{1}{2}\theta\)

Step 4 — use the tangent-radius right angle. OP is the radius drawn to the point of contact P, so by Theorem 10.1

\(\angle OPT = 90^{\circ}\)

Step 5 — subtract to get \(\angle OPQ\). The ray PQ lies inside \(\angle OPT\), so \(\angle OPQ\) is what is left of the right angle after \(\angle TPQ\) is taken away:

\(\angle OPQ = \angle OPT - \angle TPQ\)

\(\angle OPQ = 90^{\circ}-\left(90^{\circ}-\tfrac{1}{2}\theta\right)\)

\(\angle OPQ = \tfrac{1}{2}\theta = \tfrac{1}{2}\angle PTQ\)

Step 6 — read the result off. Since \(\angle OPQ = \dfrac{1}{2}\theta\) and \(\theta = \angle PTQ\), doubling both sides gives

\(\angle PTQ = 2\,\angle OPQ\)

which is what had to be proved.

Example 3

PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.

Fig. 10.10 — PQ is a chord of length 8 cm of a circle of radius 5 cm; the tangents at P and at Q meet at T, and OT cuts PQ at R

Solution

Step 1 — set the figure up. Join OT and let it cut the chord PQ at the point R (Fig. 10.10). By Theorem 10.2, TP = TQ, so △TPQ is isosceles. Also, the two right triangles OPT and OQT are congruent (OP = OQ are radii, OT is common, and both have a right angle at the point of contact), which gives \(\angle PTO = \angle QTO\) — so TO is the bisector of \(\angle PTQ\).

Step 2 — the bisector of the apex angle of an isosceles triangle is also its perpendicular bisector. Applying that to △TPQ with the bisector TR:

\(OT \perp PQ\)

\(PR = RQ = \tfrac{1}{2}(8) = 4\text{ cm}\)

Step 3 — find OR. △OPR is right-angled at R, with OP = 5 cm as its hypotenuse and PR = 4 cm:

\(OR^{2} = OP^{2}-PR^{2}\)

\(OR^{2} = 5^{2}-4^{2} = 25-16 = 9\)

\(OR = 3\text{ cm}\)

Step 4 — find a pair of similar triangles. Look at the two right triangles TRP (right angle at R) and PRO (right angle at R). In △TRP the two acute angles add to \(90^{\circ}\), and at the point of contact P the angle \(\angle OPT\) is \(90^{\circ}\), so

\(\angle TPR + \angle RPO = 90^{\circ} = \angle TPR + \angle PTR\)

Cancelling \(\angle TPR\) from both sides,

\(\angle RPO = \angle PTR\)

So △TRP and △PRO have a right angle each and one more pair of equal angles, and are therefore similar by the AA criterion:

\(\triangle TRP \sim \triangle PRO\ \text{(AA)}\)

Step 5 — use the ratio of corresponding sides.

\(\dfrac{TP}{PO} = \dfrac{RP}{RO}\)

\(\dfrac{TP}{5} = \dfrac{4}{3}\)

\(TP = \dfrac{20}{3}\text{ cm}\)

So TP = \(\dfrac{20}{3}\) cm, that is \(6\dfrac{2}{3}\) cm.

Second method — NCERT's own note, using the Pythagoras Theorem twice. Let TP = \(x\) and TR = \(y\). From the right triangle PRT (right angle at R) and the right triangle OPT (right angle at P):

\(x^{2} = y^{2} + 16 \ \text{(right }\triangle PRT)\)

\(x^{2} + 5^{2} = (y+3)^{2}\ \text{(right }\triangle OPT)\)

Subtracting the first equation from the second removes \(x^{2}\) and \(y^{2}\) together:

\(5^{2} = (y+3)^{2} - \left(y^{2}+16\right)\)

\(25 = 6y - 7\)

\(y = \dfrac{32}{6} = \dfrac{16}{3}\)

Putting this back into the first equation,

\(x^{2} = \left(\dfrac{16}{3}\right)^{2} + 16\)

\(x^{2} = \dfrac{256}{9}+\dfrac{144}{9} = \dfrac{400}{9}\)

\(x = \dfrac{20}{3}\text{ cm}\)

Both methods give the same answer, TP = \(\dfrac{20}{3}\) cm.

Exercise 10.2

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:(A) 7 cm(B) 12 cm(C) 15 cm(D) 24.5 cm

Exercise 10.2 Q1 — the tangent from Q touches the circle at P, with PQ = 24 cm and OQ = 25 cm; ∠OPQ = 90°

Answer

Step 1 — draw and label. Let O be the centre, P the point of contact of the tangent from Q, and \(r\) the radius. Then

\(PQ = 24\text{ cm},\ OQ = 25\text{ cm},\ OP = r\)

Step 2 — find the right angle. OP is the radius drawn to the point of contact and PQ is the tangent there, so by Theorem 10.1

\(\angle OPQ = 90^{\circ}\)

△OPQ is right-angled at P, and OQ — the longest side — is the hypotenuse.

Step 3 — apply the Pythagoras Theorem.

\(OQ^{2} = OP^{2} + PQ^{2}\)

\(25^{2} = r^{2} + 24^{2}\)

\(625 = r^{2} + 576\)

\(r^{2} = 625 - 576 = 49\)

\(r = 7\text{ cm}\)

So the correct option is (A) 7 cm.

Justification, and a check. The radius must be shorter than OQ = 25 cm (the point Q lies outside the circle), which already rules out (D) 24.5 cm as implausibly close and (C) 15 cm and (B) 12 cm can be tested directly: \(15^{2}+24^{2}=801\ne 625\) and \(12^{2}+24^{2}=720\ne 625\), while \(7^{2}+24^{2}=49+576=625=25^{2}\). The numbers 7, 24, 25 form a Pythagorean triple, which is exactly why they were chosen.

In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that \(\angle POQ = 110^{\circ}\), then \(\angle PTQ\) is equal to:(A) \(60^{\circ}\)(B) \(70^{\circ}\)(C) \(80^{\circ}\)(D) \(90^{\circ}\)

Fig. 10.11 — TP and TQ are tangents to the circle with centre O, and ∠POQ = 110°

Answer

Step 1 — spot the quadrilateral. The four points O, P, T and Q form the quadrilateral OPTQ. Its four angles are \(\angle POQ\) at the centre, \(\angle OPT\) and \(\angle OQT\) at the two points of contact, and \(\angle PTQ\) at the external point.

Step 2 — two of those angles are right angles. TP touches the circle at P and OP is the radius to that point of contact, and the same is true at Q. So by Theorem 10.1,

\(\angle OPT = \angle OQT = 90^{\circ}\)

Step 3 — use the angle sum of a quadrilateral. The four angles of any quadrilateral add up to \(360^{\circ}\):

\(\angle POQ + \angle OPT + \angle PTQ + \angle OQT = 360^{\circ}\)

\(110^{\circ} + 90^{\circ} + \angle PTQ + 90^{\circ} = 360^{\circ}\)

\(\angle PTQ = 360^{\circ} - 290^{\circ}\)

\(\angle PTQ = 70^{\circ}\)

So the correct option is (B) \(70^{\circ}\).

A useful shortcut worth remembering. Because the two angles at the points of contact are always \(90^{\circ}\) each, they use up \(180^{\circ}\) of the \(360^{\circ}\), leaving \(\angle PTQ + \angle POQ = 180^{\circ}\) — the two are always supplementary. That is exactly the general result proved in Q10 below, and it gives \(180^{\circ}-110^{\circ}=70^{\circ}\) in one line.

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of \(80^{\circ}\), then \(\angle POA\) is equal to:(A) \(50^{\circ}\)(B) \(60^{\circ}\)(C) \(70^{\circ}\)(D) \(80^{\circ}\)

Exercise 10.2 Q3 — the tangents PA and PB are inclined to each other at 80°; OA and OB are radii, so ∠OAP = ∠OBP = 90°

Answer

Step 1 — what is given. PA and PB are the two tangents from the external point P, touching the circle at A and B, and

\(\angle APB = 80^{\circ}\)

Step 2 — OP bisects the angle between the tangents. Compare the two triangles OAP and OBP:

\(OA = OB\ \text{(radii)}\)

\(OP = OP\ \text{(common)}\)

\(\angle OAP = \angle OBP = 90^{\circ}\)

Both are right-angled at the point of contact (Theorem 10.1), so by the RHS rule

\(\triangle OAP \cong \triangle OBP\ \text{(RHS)}\)

\(\angle APO = \angle BPO\ \text{(CPCT)}\)

Therefore OP bisects \(\angle APB\), and

\(\angle APO = \tfrac{1}{2}\angle APB = \tfrac{1}{2}\left(80^{\circ}\right) = 40^{\circ}\)

Step 3 — work inside the right triangle OAP. OA is the radius drawn to the point of contact A, so \(\angle OAP = 90^{\circ}\). Using the angle sum of △OAP,

\(\angle POA + \angle OAP + \angle APO = 180^{\circ}\)

\(\angle POA + 90^{\circ} + 40^{\circ} = 180^{\circ}\)

\(\angle POA = 50^{\circ}\)

So the correct option is (A) \(50^{\circ}\).

Justification in one line. In the right triangle OAP the two acute angles must add to \(90^{\circ}\), and one of them is half the angle between the tangents. So \(\angle POA = 90^{\circ}-\dfrac{80^{\circ}}{2}=50^{\circ}\).

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Exercise 10.2 Q4 — AB is a diameter, PQ is the tangent at A and RS the tangent at B; ∠PAB and ∠ABS are alternate angles made by the transversal AB

Answer

What is given. A circle with centre O and a diameter AB. PQ is the tangent to the circle at A, and RS is the tangent at B.

What is to be proved. PQ ∥ RS.

Step 1 — the tangent at A is perpendicular to the radius OA. By Theorem 10.1 applied at the point of contact A,

\(OA \perp PQ\)

Since B lies on the line OA (AB is a diameter, so O is the midpoint of AB and A, O, B are collinear), the ray AB points along AO. Hence

\(\angle PAB = 90^{\circ}\)

Step 2 — the same at the other end. By Theorem 10.1 applied at B, OB ⊥ RS, and by the same collinearity

\(\angle ABS = 90^{\circ}\)

Step 3 — treat AB as a transversal. The line AB cuts the two lines PQ and RS. Now \(\angle PAB\) and \(\angle ABS\) lie on opposite sides of the transversal AB and between the two lines, so they are a pair of alternate interior angles. From Steps 1 and 2,

\(\angle PAB = \angle ABS = 90^{\circ}\)

Step 4 — conclude. When a transversal cuts two lines so that a pair of alternate interior angles are equal, the two lines are parallel. Therefore

\(PQ \parallel RS\)

That is, the tangents drawn at the two ends of a diameter are parallel. (The pair \(\angle QAB\) and \(\angle ABR\) gives exactly the same conclusion, being alternate angles on the other side of the transversal.)

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Exercise 10.2 Q5 — XY is the tangent at P; the perpendicular to XY at P is drawn, and it is shown that this line has to be the line PO

Answer

What is given. A circle with centre O, and a tangent XY touching it at the point P.

What is to be proved. The line drawn perpendicular to XY at P passes through O.

Step 1 — we already know one perpendicular at P. XY is a tangent at P and OP is the radius drawn to the point of contact, so by Theorem 10.1

\(OP \perp XY\)

So the line PO is a line through P perpendicular to XY.

Step 2 — suppose the required perpendicular were a different line. Let PZ be the perpendicular to XY at P, and suppose, if possible, that PZ does not pass through O. Then PZ and PO are two different lines, and by Step 1 and by the definition of PZ,

\(\angle OPX = 90^{\circ}\ \text{and}\ \angle ZPX = 90^{\circ}\)

Step 3 — find the contradiction. That gives two distinct lines through the same point P, both perpendicular to the same line XY. But through a given point on a line, one and only one perpendicular to that line can be drawn. So the assumption in Step 2 is impossible.

Step 4 — conclude. The two lines must therefore be the same line:

\(PZ\ \text{and}\ PO\ \text{are the same line}\)

and since PO passes through the centre O, so does the perpendicular at P. Hence the perpendicular at the point of contact to a tangent passes through the centre.

Why this is worth stating separately. Theorem 10.1 runs “tangent ⇒ perpendicular to the radius”. This question runs the implication the other way, and that converse form is what lets you construct a tangent: draw the radius, raise the perpendicular at its outer end, and the line you get touches the circle there. That is exactly how the tangent in Exercise 10.1 Q4 above was drawn.

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Exercise 10.2 Q6 — the tangent from A touches the circle at B, with AB = 4 cm and OA = 5 cm; ∠OBA = 90°

Answer

Step 1 — name the points. Let O be the centre and B the point of contact of the tangent drawn from A. Then

\(OA = 5\text{ cm},\ AB = 4\text{ cm},\ OB = r\)

Step 2 — get the right angle. OB is the radius drawn to the point of contact B and AB is the tangent there, so by Theorem 10.1

\(\angle OBA = 90^{\circ}\)

So △OBA is right-angled at B, with OA as its hypotenuse.

Step 3 — apply the Pythagoras Theorem.

\(OA^{2} = OB^{2} + AB^{2}\)

\(5^{2} = r^{2} + 4^{2}\)

\(25 = r^{2} + 16\)

\(r^{2} = 25 - 16 = 9\)

\(r = 3\text{ cm}\)

Hence the radius of the circle is 3 cm.

Check. 3, 4, 5 is the smallest Pythagorean triple, and the radius (3 cm) is indeed less than the distance of A from the centre (5 cm) — as it must be, because A lies outside the circle.

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Exercise 10.2 Q7 — concentric circles of radii 5 cm and 3 cm; the chord AB of the larger circle touches the smaller circle at P, so OP ⊥ AB and P is the midpoint of AB

Answer

Step 1 — draw and label. Let O be the common centre, let the larger circle have radius 5 cm and the smaller one radius 3 cm, and let AB be the chord of the larger circle that touches the smaller circle at P. Join OP and OA. Then

\(OA = 5\text{ cm},\ OP = 3\text{ cm}\)

Step 2 — the right angle at P. AB touches the smaller circle at P and OP is the radius of the smaller circle drawn to that point of contact, so by Theorem 10.1

\(OP \perp AB\)

Step 3 — P is the midpoint of AB. This is exactly Example 1 above: OP is the perpendicular from the centre of the larger circle onto its chord AB, and the perpendicular from the centre bisects the chord. So

\(AP = PB = \tfrac{1}{2}AB\)

Step 4 — solve the right triangle OPA. △OPA is right-angled at P, with hypotenuse OA = 5 cm and one leg OP = 3 cm:

\(AP^{2} = OA^{2} - OP^{2}\)

\(AP^{2} = 5^{2}-3^{2} = 25-9 = 16\)

\(AP = 4\text{ cm}\)

Step 5 — double it. Since P is the midpoint of AB,

\(AB = 2 \times AP = 2 \times 4 = 8\text{ cm}\)

Hence the required chord is 8 cm long.

A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that \(AB + CD = AD + BC\).

Fig. 10.12 — the quadrilateral ABCD circumscribes the circle, touching it at P on AB, Q on BC, R on CD and S on DA

Answer

What is given. The quadrilateral ABCD circumscribes a circle, touching it at P on AB, at Q on BC, at R on CD and at S on DA (Fig. 10.12).

Step 1 — four pairs of equal tangent lengths. Each vertex of the quadrilateral is an external point from which two tangents are drawn, so Theorem 10.2 applies four times over:

\(AP = AS\ \text{(tangents from }A)\)

\(BP = BQ\ \text{(tangents from }B)\)

\(CR = CQ\ \text{(tangents from }C)\)

\(DR = DS\ \text{(tangents from }D)\)

Step 2 — add all four equations. Adding the left sides and the right sides,

\(AP + BP + CR + DR = AS + BQ + CQ + DS\)

Step 3 — regroup the terms into whole sides. The points of contact lie on the sides, so each side is the sum of the two tangent lengths along it. Grouping the left side as (AP + PB) and (CR + RD), and the right side as (AS + SD) and (BQ + QC):

\((AP + PB) + (CR + RD) = (AS + SD) + (BQ + QC)\)

\(AB + CD = AD + BC\)

Step 4 — read the sides off. Since P lies on AB, R on CD, S on DA and Q on BC,

\(AP + PB = AB,\ CR + RD = CD\)

\(AS + SD = AD,\ BQ + QC = BC\)

Substituting these into Step 3 gives

\(AB + CD = AD + BC\)

which is what had to be proved.

What the result says. In any quadrilateral that can be drawn around a circle, the two pairs of opposite sides have the same total. This is used directly in Q11 below to prove that a circumscribing parallelogram has to be a rhombus.

In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that \(\angle AOB = 90^{\circ}\).

Fig. 10.13 — XY and X′Y′ are parallel tangents touching at P and Q; a third tangent AB touches at C, meeting XY at A and X′Y′ at B

Answer

What is given. XY and X′Y′ are parallel tangents touching the circle at P and Q; AB is a third tangent touching at C, meeting XY at A and X′Y′ at B. Join OP, OQ, OC, OA and OB.

Step 1 — OA bisects the angle at A. From the external point A two tangents are drawn: AP (part of XY) and AC (part of AB). Compare △OPA and △OCA, using Theorem 10.2 for the equal tangent lengths:

\(OP = OC\ \text{(radii of the same circle)}\)

\(AP = AC\ \text{(tangents from }A)\)

\(OA = OA\ \text{(common side)}\)

So by the SSS rule △OPA ≅ △OCA, and therefore

\(\angle OAP = \angle OAC\ \text{(CPCT)}\)

That is, OA bisects \(\angle PAC\), which is the angle \(\angle PAB\).

Step 2 — OB bisects the angle at B. Exactly the same argument at the external point B, using the tangents BQ and BC, gives △OQB ≅ △OCB and hence

\(\angle OBQ = \angle OBC\)

Step 3 — use the parallel tangents. XY ∥ X′Y′ and AB is a transversal cutting them, so \(\angle PAB\) and \(\angle QBA\) are co-interior (allied) angles on the same side of the transversal, and therefore add up to two right angles:

\(\angle PAB + \angle QBA = 180^{\circ}\)

Step 4 — halve that equation. Dividing both sides by 2 and using Steps 1 and 2,

\(\tfrac{1}{2}\angle PAB + \tfrac{1}{2}\angle QBA = 90^{\circ}\)

\(\angle OAB + \angle OBA = 90^{\circ}\)

Step 5 — finish inside triangle AOB. The angles of △AOB add to \(180^{\circ}\):

\(\angle OAB + \angle OBA + \angle AOB = 180^{\circ}\)

\(90^{\circ} + \angle AOB = 180^{\circ}\)

\(\angle AOB = 90^{\circ}\)

Hence \(\angle AOB = 90^{\circ}\), as required.

Q10

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Exercise 10.2 Q10 — PA and PB are the two tangents from the external point P, touching at A and B; OAPB is a quadrilateral with right angles at A and at B

Answer

What is given. A circle with centre O and an external point P from which two tangents PA and PB are drawn, touching the circle at A and B.

What is to be proved. \(\angle APB + \angle AOB = 180^{\circ}\) — the angle between the tangents and the angle subtended at the centre by the chord AB are supplementary.

Step 1 — look at the quadrilateral OAPB. Joining OA, OB, PA and PB gives the quadrilateral OAPB, whose four angles are \(\angle AOB\) at the centre, \(\angle OAP\) and \(\angle OBP\) at the two points of contact, and \(\angle APB\) at P.

Step 2 — the two angles at the points of contact are right angles. PA is a tangent at A and OA is the radius drawn to that point of contact, and the same holds at B. So by Theorem 10.1,

\(\angle OAP = \angle OBP = 90^{\circ}\)

Step 3 — use the angle sum of a quadrilateral. The four angles of OAPB add to \(360^{\circ}\):

\(\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^{\circ}\)

\(\angle AOB + 90^{\circ} + \angle APB + 90^{\circ} = 360^{\circ}\)

\(\angle AOB + \angle APB = 360^{\circ} - 180^{\circ}\)

\(\angle APB + \angle AOB = 180^{\circ}\)

Hence the angle between the two tangents and the angle subtended at the centre by the segment joining the points of contact are supplementary.

Where this gets used. This is the general result behind Q2 above: with \(\angle POQ = 110^{\circ}\) it gives \(\angle PTQ = 180^{\circ}-110^{\circ}=70^{\circ}\) in a single step.

Q11

Prove that the parallelogram circumscribing a circle is a rhombus.

Exercise 10.2 Q11 — the parallelogram ABCD circumscribes the circle, touching AB at P, BC at Q, CD at R and DA at S

Answer

What is given. A parallelogram ABCD which circumscribes a circle, touching it at P on AB, Q on BC, R on CD and S on DA.

What is to be proved. ABCD is a rhombus, that is, all four of its sides are equal.

Step 1 — what being a parallelogram gives. In a parallelogram the opposite sides are equal:

\(AB = CD\ \text{and}\ BC = AD\)

Step 2 — what circumscribing the circle gives. ABCD is a quadrilateral drawn around a circle, so the result of Q8 applies:

\(AB + CD = AD + BC\)

Step 3 — substitute Step 1 into Step 2. Replace CD by AB and AD by BC:

\(AB + AB = BC + BC\)

\(2\,AB = 2\,BC\)

\(AB = BC\)

Step 4 — chain the equalities together. Combining \(AB = BC\) from Step 3 with \(AB = CD\) and \(BC = AD\) from Step 1,

\(AB = BC = CD = DA\)

All four sides are equal, so the parallelogram ABCD is a rhombus.

The idea in one sentence. Being a parallelogram makes the two pairs of opposite sides equal; circumscribing the circle makes the two pairs have equal totals; together those force all four sides to be the same length.

Q12

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Fig. 10.14 — the triangle ABC is drawn to circumscribe a circle of radius 4 cm, whose point of contact D divides BC into CD = 6 cm and DB = 8 cm

Answer

Step 1 — name the other two points of contact. Let the circle touch BC at D, CA at E and AB at F. By Theorem 10.2 the two tangent lengths from each vertex are equal, so writing \(AE = AF = x\) cm:

\(BD = BF = 8\text{ cm}\)

\(CD = CE = 6\text{ cm}\)

\(AE = AF = x\text{ cm}\)

Therefore the three sides of the triangle are

\(BC = 8+6 = 14,\ CA = x+6,\ AB = x+8\)

Step 2 — write down the semi-perimeter.

\(s = \dfrac{AB + BC + CA}{2}\)

\(s = \dfrac{(x+8)+14+(x+6)}{2}\)

\(s = \dfrac{2x+28}{2} = x+14\)

Step 3 — the area, cut into three triangles from the centre. Joining the centre O to A, B and C splits △ABC into △OBC, △OCA and △OAB. Each has one side of the triangle as its base and the radius (4 cm) as its height, because the radius drawn to a point of contact is perpendicular to that side:

\(\text{Area} = \tfrac{1}{2}(BC)(4) + \tfrac{1}{2}(CA)(4) + \tfrac{1}{2}(AB)(4)\)

\(\text{Area} = 2\left[14 + (x+6) + (x+8)\right] = 2(2x+28)\)

\(\text{Area} = 4(x+14)\)

Step 4 — the area again, by Heron's formula. With \(s = x+14\), \(s-BC = x\), \(s-CA = 8\) and \(s-AB = 6\):

\(\text{Area} = \sqrt{s(s-BC)(s-CA)(s-AB)}\)

\(\text{Area} = \sqrt{(x+14)\,(x)\,(8)\,(6)} = \sqrt{48x(x+14)}\)

Step 5 — equate the two expressions and solve.

\(4(x+14) = \sqrt{48x(x+14)}\)

Squaring both sides,

\(16(x+14)^{2} = 48x(x+14)\)

Since \(x+14\) is a length and cannot be zero, divide both sides by \(16(x+14)\):

\(16(x+14) = 48x\)

\(x + 14 = 3x\)

\(x = 7\)

Step 6 — write down the sides.

\(AB = x + 8 = 7 + 8 = 15\text{ cm}\)

\(AC = x + 6 = 7 + 6 = 13\text{ cm}\)

So AB = 15 cm and AC = 13 cm.

Check. The finished triangle has sides 13 cm, 14 cm and 15 cm, so \(s = \dfrac{13+14+15}{2} = 21\) and Heron's formula gives an area of \(\sqrt{21 \times 7 \times 8 \times 6} = \sqrt{7056} = 84\) cm². The inradius is then \(\dfrac{\text{area}}{s} = \dfrac{84}{21} = 4\) cm, matching the radius given in the question, and the tangent lengths come out as \(s - AC = 21-13 = 8\) cm and \(s - AB = 21-15 = 6\) cm — exactly the BD and DC the question started from.

Q13

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Exercise 10.2 Q13 — the quadrilateral ABCD circumscribes the circle with centre O; joining O to each vertex and to each point of contact splits the full angle at O into eight parts

Answer

What is given. A quadrilateral ABCD circumscribing a circle with centre O, touching it at P on AB, Q on BC, R on CD and S on DA. Join O to each of A, B, C, D and to each of P, Q, R, S.

What is to be proved. \(\angle AOB + \angle COD = 180^{\circ}\) and \(\angle BOC + \angle AOD = 180^{\circ}\).

Step 1 — name the eight angles at O. Going once round the centre, the eight rays OA, OP, OB, OQ, OC, OR, OD, OS cut the complete angle at O into eight parts. Name them in that order:

\(\angle 1 = \angle AOP,\ \angle 2 = \angle POB,\ \angle 3 = \angle BOQ,\ \angle 4 = \angle QOC\)

\(\angle 5 = \angle COR,\ \angle 6 = \angle ROD,\ \angle 7 = \angle DOS,\ \angle 8 = \angle SOA\)

Step 2 — pair them up using congruent triangles. At the vertex A the two tangents are AP and AS. Compare △OAP and △OAS, using Theorem 10.2 for the equal tangent lengths:

\(OP = OS\ \text{(radii of the same circle)}\)

\(AP = AS\ \text{(tangents from }A)\)

\(OA = OA\ \text{(common side)}\)

So △OAP ≅ △OAS by SSS, and therefore \(\angle AOP = \angle AOS\), that is

\(\angle 1 = \angle 8\)

Repeating the identical argument at B, at C and at D,

\(\angle 2 = \angle 3,\ \angle 4 = \angle 5,\ \angle 6 = \angle 7\)

Step 3 — the eight angles fill the complete angle at O.

\(\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8 = 360^{\circ}\)

Step 4 — replace each angle by its partner. Using \(\angle 8 = \angle 1\), \(\angle 3 = \angle 2\), \(\angle 4 = \angle 5\) and \(\angle 7 = \angle 6\), the sum in Step 3 becomes

\(2\angle 1 + 2\angle 2 + 2\angle 5 + 2\angle 6 = 360^{\circ}\)

\(\angle 1 + \angle 2 + \angle 5 + \angle 6 = 180^{\circ}\)

Step 5 — recognise the two angles we want. The ray OP lies inside \(\angle AOB\) and the ray OR lies inside \(\angle COD\), so those two angles split as

\(\angle AOB = \angle 1 + \angle 2,\ \angle COD = \angle 5 + \angle 6\)

Adding them and using Step 4,

\(\angle AOB + \angle COD = 180^{\circ}\)

Step 6 — the other pair. The four angles at O around a point add to \(360^{\circ}\), so

\(\angle AOB + \angle BOC + \angle COD + \angle AOD = 360^{\circ}\)

\(\angle BOC + \angle AOD = 360^{\circ} - 180^{\circ} = 180^{\circ}\)

Hence both pairs of opposite sides subtend supplementary angles at the centre, as required.

Chapter 9: Some Applications of Trigonometry

All Chapters

Chapter 11: Areas Related to Circles

Circles

Key Theorems Used on This Page

Exercise 10.1

Worked Examples (Examples 1–3)

Exercise 10.2

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