NCERT Exemplar Solutions Class 10 Maths Circles — Exercises 9.3 and 9.4

NCERT Exemplar Problems, Class 10 Mathematics — Circles, Exercises 9.3 and 9.4. All 22 questions solved.

Twenty-two questions, and most of them are one idea applied with care: the two tangents from a point are equal. That single fact, used four or six times in the same figure, proves every result about circumscribed triangles, quadrilaterals and hexagons in this exercise.

The technique is always the same, and it is worth stating as a procedure:

1. Name the tangent lengths, not the sides. If the incircle of a triangle touches the sides at $D$, $E$, $F$, call the two equal tangents from $A$ both $x$, from $B$ both $y$, from $C$ both $z$. Every side is then a sum of two of them: $a = y + z$, $b = z + x$, $c = x + y$.
2. Add up whatever the question asks for, in terms of $x$, $y$, $z$.
3. Watch the cancellation. Questions 1 and 2 of Exercise 9.4 both come out in one line this way.

The perimeter results (Exercise 9.4, Questions 3 and 14) use the same idea in its most striking form: a whole path of tangent segments collapses into $PA + PB$, so the perimeter is twice the tangent length no matter where the third tangent is drawn.

A note on the current syllabus. The rationalised Class 10 chapter carries just two theorems — the tangent is perpendicular to the radius at the point of contact, and the two tangents from an external point are equal — and the tangent–chord theorem is not among them. Three questions here are usually solved with it, and each is done instead by joining the centre or by using the fact that a radius perpendicular to a chord bisects it. Those are the tools your textbook actually gives you.

Key insight. Whenever two common tangents to two circles appear, find the point where they meet. Every result about common tangents in this exercise — Questions 5, 7 of Exercise 9.3 and Question 10 of Exercise 9.4 — follows from applying “equal tangents from a point” at that meeting point, to each circle in turn.

Exercise 9.3

Question 1

Out of the two concentric circles, the radius of the outer circle is $5$ cm and the chord $AC$ of length $8$ cm is a tangent to the inner circle. Find the radius of the inner circle.

Solution. Let the chord $AC$ touch the inner circle at $M$. Then $OM$ is a radius of the inner circle drawn to its point of contact, so $OM \perp AC$.

A radius perpendicular to a chord bisects it, so

$$AM = \frac{AC}{2} = \frac{8}{2} = 4 \text{ cm}$$

Now Pythagoras in the right-angled triangle $OMA$, where $OA = 5$ is the radius of the outer circle:

$$OM = \sqrt{OA^2 – AM^2} = \sqrt{25 – 16} = 3 \text{ cm}$$

This is Question 1 of Exercise 9.1 run in reverse: there the radii were given and the chord was asked for, here the chord is given and one radius is asked for. The halving step appears in both, and in opposite directions — halve the chord to get into the triangle, double the half-chord to get out of it.

Question 2

Two tangents $PQ$ and $PR$ are drawn from an external point to a circle with centre $O$. Prove that $QORP$ is a cyclic quadrilateral.

Solution. A quadrilateral is cyclic exactly when a pair of opposite angles are supplementary, so name the angles of $QORP$ and look for such a pair.

$OQ$ is the radius to the point of contact $Q$ and $PQ$ is the tangent there, so by Theorem 1

$$\angle OQP = 90^\circ$$

Likewise at $R$:

$$\angle ORP = 90^\circ$$

In the quadrilateral $QORP$ the vertices are visited in that order, so $\angle Q$ and $\angle R$ are at opposite corners — $Q$ is the first vertex and $R$ the third. Their sum is

$$90^\circ + 90^\circ = 180^\circ$$

Since one pair of opposite angles is supplementary, $QORP$ is cyclic.

The circle through all four points is easy to name: with $\angle OQP = \angle ORP = 90^\circ$, both $Q$ and $R$ lie on the circle having $OP$ as diameter, by the converse of the angle in a semicircle. So $QORP$ is inscribed in the circle on $OP$ as diameter, whose centre is the mid-point of $OP$.

Question 3

If from an external point $B$ of a circle with centre $O$, two tangents $BC$ and $BD$ are drawn such that $\angle DBC = 120^\circ$, prove that $BC + BD = BO$, i.e., $BO = 2BC$.

Solution. The line $BO$ bisects the angle between the tangents, because triangles $OCB$ and $ODB$ are congruent ($OC = OD$ radii, $BC = BD$ tangents, $OB$ common). So

$$\angle OBC = \frac{120^\circ}{2} = 60^\circ$$

Now work in triangle $OCB$, which is right angled at $C$ since $OC$ is the radius to the point of contact. Relative to the $60^\circ$ angle at $B$, the tangent $BC$ is adjacent and $BO$ is the hypotenuse:

$$\cos 60^\circ = \frac{BC}{BO} \implies \frac12 = \frac{BC}{BO} \implies BO = 2BC$$

And since the two tangents are equal, $BC + BD = BC + BC = 2BC = BO$, which is the other form of the same statement.

Why $120^\circ$ makes it work. The half-angle is $60^\circ$, and $\cos 60^\circ = \frac12$ is the one value that turns the relation into a clean doubling. Any other angle gives $BO = BC\sec(\theta/2)$ with an awkward constant. The tangent length here is $BC = BO\cos 60^\circ$ and the radius is $OC = BO\sin 60^\circ = \frac{\sqrt3}{2}BO$, so the tangent is shorter than the radius — an instance of the situation Exercise 9.2, Question 2 was about.

Question 4

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Solution. Let the two lines meet at $P$ and let a circle of centre $O$ touch them at $A$ and $B$.

The distances from $O$ to the two lines are equal. By Theorem 1, $OA \perp PA$ and $OB \perp PB$, so $OA$ and $OB$ are precisely the perpendicular distances from $O$ to the two lines. Both are radii of the same circle, so

$$OA = OB = r$$

A point equidistant from two intersecting lines lies on an angle bisector. That is the characterisation of the bisector, and it is what finishes the proof.

If you would rather see the congruence explicitly: in triangles $OAP$ and $OBP$ we have $\angle OAP = \angle OBP = 90^\circ$, $OA = OB$ (radii) and $OP$ common, so the triangles are congruent by RHS. Corresponding angles then give

$$\angle OPA = \angle OPB$$

so $PO$ divides the angle at $P$ into two equal parts — it is the bisector.

This is the fact that makes an incircle exist. Each of a triangle’s three angle bisectors is the locus of centres of circles touching two of its sides; all three meet at one point, and that point is equidistant from all three sides, so a single circle touches all three. The incentre of Exercise 9.4’s Question 2 is exactly this.

Question 5

In Fig. 9.13, $AB$ and $CD$ are common tangents to two circles of unequal radii. Prove that $AB = CD$.

Solution. The two circles have unequal radii, so the two common tangents are not parallel: extended, they meet at a point. Call it $P$, as in the figure above (the book does not draw it, and it is the whole proof).

Now apply “equal tangents from a point” twice from $P$ — once to each circle.

To the larger circle. $PA$ and $PC$ are both tangents from $P$, so

$$PA = PC$$

To the smaller circle. $PB$ and $PD$ are both tangents from $P$, so

$$PB = PD$$

Subtract. $A$ and $B$ lie on the same tangent line with $B$ between $A$ and $P$, and likewise $D$ between $C$ and $P$. So

$$AB = PA – PB \qquad\text{and}\qquad CD = PC – PD$$

Both differences are of equal quantities, so

$$AB = PA – PB = PC – PD = CD$$

as required.

The unequal radii are essential to this argument, because they are what guarantees $P$ exists. That is exactly what the next question is about.

Question 6

In Question 5 above, if radii of the two circles are equal, prove that $AB = CD$.

Solution. The same conclusion, but the proof of Question 5 breaks down and has to be replaced — which is why this appears as a separate question.

With equal radii there is no point $P$: the two common tangents are parallel, so they never meet and “tangents from $P$” means nothing. The argument must come from somewhere else.

Let the circles have centres $O$ and $O’$ and equal radius $r$, with $AB$ touching them at $A$ and $B$.

$OABO’$ is a rectangle. By Theorem 1, $OA \perp AB$ and $O’B \perp AB$. So $OA$ and $O’B$ are both perpendicular to the same line, hence parallel to each other; and they are equal, both being radii of length $r$.

A quadrilateral with one pair of sides both equal and parallel is a parallelogram, so $OABO’$ is a parallelogram — and since its angles at $A$ and $B$ are right angles, it is a rectangle. Therefore its opposite sides are equal:

$$AB = OO’$$

The same argument on the other tangent gives $OCDO’$ a rectangle and

$$CD = OO’$$

Both tangents equal the distance between the centres, so

$$AB = CD = OO’$$

That is a stronger result than the question asks for, and a more useful one: with equal circles, each common external tangent has exactly the length of the line of centres.

Question 7

In Fig. 9.14, common tangents $AB$ and $CD$ to two circles intersect at $E$. Prove that $AB = CD$.

Solution. These are the internal common tangents — they cross between the circles rather than running along the outside — and the meeting point $E$ is given rather than constructed. That makes the proof shorter than Question 5’s.

Apply “equal tangents from a point” twice from $E$.

To the first circle, touched at $A$ and $C$:

$$EA = EC$$

To the second circle, touched at $B$ and $D$:

$$EB = ED$$

Add, rather than subtract. Here $E$ lies between $A$ and $B$ on one tangent, and between $C$ and $D$ on the other — that is what “intersect at $E$” means for internal tangents. So

$$AB = EA + EB \qquad\text{and}\qquad CD = EC + ED$$

and therefore

$$AB = EA + EB = EC + ED = CD$$

Compare with Question 5. The two proofs use exactly the same fact and differ only in whether the meeting point lies inside or outside the segment — giving a sum here and a difference there. Getting that the wrong way round is the only real hazard in either, and the figure settles it: look at whether $E$ (or $P$) is between the two points of contact.

Question 8

A chord $PQ$ of a circle is parallel to the tangent drawn at a point $R$ of the circle. Prove that $R$ bisects the arc $PRQ$.

Solution. Let $O$ be the centre and let the tangent at $R$ be the line $\ell$, with $PQ \parallel \ell$.

Step 1 — $OR \perp PQ$. By Theorem 1, $OR \perp \ell$. Since $PQ \parallel \ell$, the same perpendicular applies:

$$OR \perp PQ$$

Step 2 — $OR$ bisects the chord. A radius perpendicular to a chord passes through its mid-point. So the line $OR$, extended, meets $PQ$ at its mid-point $M$, and $PM = MQ$.

Step 3 — the arcs are equal. Triangles $OPM$ and $OQM$ are congruent (SSS: $OP = OQ$ radii, $PM = MQ$, $OM$ common), so the angles at $O$ are equal:

$$\angle POR = \angle QOR$$

Equal angles at the centre stand on equal arcs. So arc $PR$ and arc $RQ$ are equal, which is precisely what “$R$ bisects arc $PRQ$” means.

The result is symmetry, stated in arcs. The line $OR$ is an axis of symmetry of the whole figure: reflecting in it fixes the circle, fixes $R$ and the tangent, and swaps $P$ with $Q$. Everything in the proof is that one observation written out.

Published solutions usually run this through the tangent–chord theorem, showing $\angle QPR = \angle PQR$ and hence $PR = QR$. That theorem is not in the current Class 10 textbook; the perpendicular-from-the-centre route uses only what is.

Exercise 9.4

Question 9

If a hexagon $ABCDEF$ circumscribes a circle, prove that $AB + CD + EF = BC + DE + FA$.

(Exercise 9.4, Question 1.)

Solution. The incircle touches each of the six sides once, so each vertex is the source of exactly two tangent segments — and those two are equal.

Call the tangent length from each vertex by the lower-case letter of that vertex:

$$a \text{ from } A, \quad b \text{ from } B, \quad c \text{ from } C, \quad d \text{ from } D, \quad e \text{ from } E, \quad f \text{ from } F$$

Every side is split by its point of contact into the tangent segments from its two endpoints:

$$AB = a + b, \quad BC = b + c, \quad CD = c + d, \quad DE = d + e, \quad EF = e + f, \quad FA = f + a$$

Now add the two groups the question names.

$$AB + CD + EF = (a+b) + (c+d) + (e+f) = a + b + c + d + e + f$$

$$BC + DE + FA = (b+c) + (d+e) + (f+a) = a + b + c + d + e + f$$

The two totals are the same sum, so they are equal.

The reason it works is that the two groups alternate. Taking every other side picks up each vertex exactly once, so both groups collect all six tangent lengths. The same argument proves the corresponding result for a circumscribed quadrilateral — $AB + CD = BC + DA$ — and for any circumscribed polygon with an even number of sides.

Question 10

Let $s$ denote the semi-perimeter of a triangle $ABC$ in which $BC = a$, $CA = b$, $AB = c$. If a circle touches the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively, prove that $BD = s – b$.

(Exercise 9.4, Question 2.)

Solution. Name the tangent lengths from the three vertices:

$$BD = BF = x, \qquad CD = CE = y, \qquad AE = AF = z$$

Each is equal to its partner because the two tangents from a vertex are equal.

Now express the three sides:

$$a = BC = BD + DC = x + y$$

$$b = CA = CE + EA = y + z$$

$$c = AB = AF + FB = z + x$$

Add all three:

$$a + b + c = 2(x + y + z) \implies x + y + z = \frac{a+b+c}{2} = s$$

So the sum of the three tangent lengths is exactly the semi-perimeter — which is the fact worth remembering from this question.

Finally, subtract $b$ from that:

$$s – b = (x + y + z) – (y + z) = x = BD$$

as required.

By the same argument, $CD = s – c$ and $AE = s – a$: the tangent length from a vertex is the semi-perimeter minus the opposite side. That is the form in which the result is used, and it reappears in the standard incircle radius formula $r = \frac{\text{area}}{s}$.

Question 11

From an external point $P$, two tangents, $PA$ and $PB$ are drawn to a circle with centre $O$. At one point $E$ on the circle tangent is drawn which intersects $PA$ and $PB$ at $C$ and $D$, respectively. If $PA = 10$ cm, find the perimeter of the triangle $PCD$.

(Exercise 9.4, Question 3.)

Solution. Write the perimeter out and then break the side $CD$ at the point where the third tangent touches:

$$\text{perimeter} = PC + CD + DP = PC + \underbrace{CE + ED}_{= \,CD} + DP$$

Now use equal tangents at $C$ and at $D$.

From $C$, the tangents to the circle are $CA$ (along the line $PA$) and $CE$, so

$$CE = CA$$

From $D$, the tangents are $DB$ and $DE$, so

$$ED = DB$$

Substituting:

$$\text{perimeter} = PC + CA + DB + DP = (PC + CA) + (DP + DB) = PA + PB$$

because $C$ lies on $PA$ between $P$ and $A$, and $D$ lies on $PB$ between $P$ and $B$.

Finally $PB = PA = 10$ cm (equal tangents from $P$), so

$$\text{perimeter} = 10 + 10 = 20 \text{ cm}$$

Nothing depended on where $E$ was. Slide the point of contact anywhere along the arc between $A$ and $B$ and the triangle $PCD$ changes shape completely, but its perimeter stays at $20$ cm. That is the striking part of the result, and it is worth stating: the perimeter is always twice the tangent length.

Question 12

If $AB$ is a chord of a circle with centre $O$, $AOC$ is a diameter and $AT$ is the tangent at $A$ as shown in Fig. 9.17. Prove that $\angle BAT = \angle ACB$.

(Exercise 9.4, Question 4.)

Solution. This is the general statement behind Question 3 of Exercise 9.1, where the same configuration had $\angle ACB = 50^\circ$ and the answer came out as $50^\circ$ too. Here is why it must.

Step 1 — the angle in a semicircle. $AOC$ is a diameter and $B$ is on the circle, so

$$\angle ABC = 90^\circ$$

Step 2 — the angle sum in triangle $ABC$.

$$\angle BAC + \angle ACB = 180^\circ – 90^\circ = 90^\circ$$

so the two acute angles of the triangle are complementary:

$$\angle BAC = 90^\circ – \angle ACB$$

Step 3 — the tangent–radius right angle. $OA$ is a radius and $AT$ is the tangent at $A$, so $\angle OAT = 90^\circ$. Since $C$ lies on the ray $AO$, the angle $\angle CAT$ is that same $90^\circ$, and $\angle BAC$ sits inside it:

$$\angle BAT = 90^\circ – \angle BAC$$

Step 4 — combine. Substituting Step 2 into Step 3:

$$\angle BAT = 90^\circ – \left(90^\circ – \angle ACB\right) = \angle ACB$$

as required.

What has just been proved is the tangent–chord theorem, in the special case where the chord’s other end is joined to a point of the circle via a diameter. The general theorem — the angle between a tangent and a chord equals the angle the chord subtends in the alternate segment — is no longer in the Class 10 textbook, but this question reconstructs the instance of it that matters, using only the angle in a semicircle and the tangent–radius right angle.

Question 13

Two circles with centres $O$ and $O’$ of radii $3$ cm and $4$ cm, respectively intersect at two points $P$ and $Q$ such that $OP$ and $O’P$ are tangents to the two circles. Find the length of the common chord $PQ$.

(Exercise 9.4, Question 5.)

Solution. The condition in the question is doing a lot of work, so unpack it first.

$OP$ is a radius of the first circle. It is also stated to be a tangent to the second circle, and it touches that circle at $P$. By Theorem 1, a tangent is perpendicular to the radius of the circle it touches at the point of contact — here that radius is $O’P$. So

$$OP \perp O’P, \qquad \text{that is} \qquad \angle OPO’ = 90^\circ$$

Find $OO’$. Triangle $OPO’$ is right angled at $P$ with legs $OP = 3$ and $O’P = 4$:

$$OO’ = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \text{ cm}$$

Find the common chord. The common chord $PQ$ is perpendicular to the line of centres $OO’$ and is bisected by it. Let $M$ be their intersection. Then $PM$ is the altitude from the right angle of triangle $OPO’$ to its hypotenuse, and the standard relation gives

$$PM = \frac{OP \times O’P}{OO’} = \frac{3 \times 4}{5} = 2.4 \text{ cm}$$

(That relation is just “area two ways”: $\frac12 \times 3 \times 4 = \frac12 \times 5 \times PM$.)

Finally the chord is twice the half-chord:

$$PQ = 2 \times 2.4 = 4.8 \text{ cm}$$

Question 14

In a right triangle $ABC$ in which $\angle B = 90^\circ$, a circle is drawn with $AB$ as diameter intersecting the hypotenuse $AC$ at $P$. Prove that the tangent to the circle at $P$ bisects $BC$.

(Exercise 9.4, Question 6.)

Solution. Let the tangent at $P$ meet $BC$ at $M$. The goal is to show $MB = MC$.

Step 1 — $BC$ is itself a tangent to the circle. $AB$ is a diameter, so $B$ is on the circle and $AB$ is the radius line there. Since $\angle B = 90^\circ$, the line $BC$ is perpendicular to $AB$ at $B$ — which is exactly the condition for $BC$ to be the tangent at $B$.

Step 2 — $MP = MB$. $M$ lies on both tangents, at $P$ and at $B$, so these are two tangents from the same external point:

$$MP = MB$$

Step 3 — $MP = MC$. Since $AB$ is a diameter and $P$ is on the circle, $\angle APB = 90^\circ$, and so $\angle BPC = 90^\circ$ as well ($A$, $P$, $C$ are collinear).

Now compare two angles at $P$. From $MP = MB$ the triangle $MPB$ is isosceles, so $\angle MPB = \angle MBP$. Therefore

$$\angle MPC = 90^\circ – \angle MPB = 90^\circ – \angle MBP$$

And in the triangle $BPC$, right angled at $P$:

$$\angle MCP = \angle BCP = 90^\circ – \angle PBC = 90^\circ – \angle MBP$$

since $\angle PBC$ and $\angle MBP$ are the same angle ($M$ lies on $BC$). So $\angle MPC = \angle MCP$, the triangle $MPC$ is isosceles, and

$$MP = MC$$

Step 4. Combining Steps 2 and 3:

$$MB = MP = MC$$

so $M$ is the mid-point of $BC$ and the tangent at $P$ bisects it.

The point $M$ turns out to be equidistant from $B$, $P$ and $C$ — it is the circumcentre of the right-angled triangle $BPC$, sitting at the mid-point of its hypotenuse, exactly as Question 15 of Exercise 7.1 would predict.

Question 15

In Fig. 9.18, tangents $PQ$ and $PR$ are drawn to a circle such that $\angle RPQ = 30^\circ$. A chord $RS$ is drawn parallel to the tangent $PQ$. Find the $\angle RQS$.

(Exercise 9.4, Question 7.)

Solution. Three steps, and the parallel chord is used twice.

Step 1 — the triangle $PQR$. The tangents from $P$ are equal, so $PQ = PR$ and the triangle is isosceles. Its base angles are

$$\angle PQR = \angle PRQ = \frac{180^\circ – 30^\circ}{2} = 75^\circ$$

Step 2 — transfer across the parallels. $RS \parallel PQ$ with $QR$ as transversal, so $\angle SRQ$ and $\angle PQR$ are alternate angles:

$$\angle SRQ = \angle PQR = 75^\circ$$

Step 3 — the triangle $QRS$ is isosceles. Let $O$ be the centre. $OQ \perp PQ$ (tangent–radius), and $RS \parallel PQ$, so $OQ \perp RS$ too. A radius perpendicular to a chord bisects it, so $OQ$ passes through the mid-point of $RS$ — making the whole figure symmetric about the line $OQ$, which swaps $R$ with $S$ and fixes $Q$.

Therefore $QR = QS$ and

$$\angle QSR = \angle QRS = 75^\circ$$

Step 4 — angle sum.

$$\angle RQS = 180^\circ – 75^\circ – 75^\circ = 30^\circ$$

The answer equals the given angle exactly, and the figure above is drawn to scale — $\angle RQS$ measures $29.97^\circ$ in it. That coincidence is not an accident of the numbers: for a tangent angle of $2\theta$ the base angles are $90^\circ – \theta$ and the apex angle comes out as $2\theta$ again, for every $\theta$.

The book’s hint suggests drawing the line through $Q$ perpendicular to $QP$ — which is the line $OQ$ of Step 3, reached without naming the centre. Published solutions instead use the tangent–chord theorem to get $\angle QSR = \angle PQR$ in one step; that theorem is not in the current Class 10 textbook.

Question 16

$AB$ is a diameter and $AC$ is a chord of a circle with centre $O$ such that $\angle BAC = 30^\circ$. The tangent at $C$ intersects extended $AB$ at a point $D$. Prove that $BC = BD$.

(Exercise 9.4, Question 8.)

Solution. This is Question 10 of Exercise 9.2 set as a proof rather than a true-or-false judgement, and the reasoning is the same. Join $O$ to $C$.

Step 1. $AB$ is a diameter, so $\angle ACB = 90^\circ$ (angle in a semicircle), and with $\angle BAC = 30^\circ$ the angle sum gives

$$\angle ABC = 180^\circ – 90^\circ – 30^\circ = 60^\circ$$

Step 2. $OB = OC$ are radii, so triangle $OBC$ is isosceles with $\angle OCB = \angle OBC = 60^\circ$. A triangle with two $60^\circ$ angles is equilateral, so all three angles are $60^\circ$ and, incidentally, $BC = OB = r$.

Step 3. $OC$ is the radius to the point of contact and $CD$ is the tangent there, so $\angle OCD = 90^\circ$. The angle $\angle OCB$ occupies $60^\circ$ of it:

$$\angle BCD = 90^\circ – 60^\circ = 30^\circ$$

Step 4. $D$ lies on $AB$ produced beyond $B$, so $\angle CBD$ is the supplement of $\angle ABC$:

$$\angle CBD = 180^\circ – 60^\circ = 120^\circ$$

and the angle sum in triangle $BCD$ gives

$$\angle BDC = 180^\circ – 120^\circ – 30^\circ = 30^\circ$$

Step 5. In triangle $BCD$, $\angle BCD = \angle BDC = 30^\circ$, so it is isosceles and the sides opposite those equal angles are equal:

$$BC = BD$$

A remark on the $30^\circ$. Step 2 only produced an equilateral triangle because $\angle ABC$ came out as exactly $60^\circ$, which needed $\angle BAC = 30^\circ$. With any other angle the result fails — the question is not a general fact about diameters and tangents but a consequence of this particular value.

Question 17

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

(Exercise 9.4, Question 9.)

Solution. This is the converse of Question 8 of Exercise 9.3, and the same symmetry proves it in the other direction.

Let $R$ be the mid-point of the arc $PQ$, let $O$ be the centre, and let $\ell$ be the tangent at $R$.

Step 1 — $OR$ is the perpendicular bisector of $PQ$. Since $R$ is the mid-point of the arc, the arcs $PR$ and $RQ$ are equal, so the central angles standing on them are equal:

$$\angle POR = \angle QOR$$

Triangles $POR$ and $QOR$ are therefore congruent (SAS: $OP = OQ$ radii, the equal angles at $O$, $OR$ common), which gives $RP = RQ$. So $R$ is equidistant from $P$ and $Q$, and so is $O$ — both lie on the perpendicular bisector of $PQ$.

Two points determine a line, so the line $OR$ is that perpendicular bisector, and in particular

$$OR \perp PQ$$

Step 2 — $OR \perp \ell$. That is Theorem 1: the tangent at $R$ is perpendicular to the radius $OR$.

Step 3. Two lines perpendicular to the same line are parallel:

$$\ell \parallel PQ$$

Notice that this is the identical three-step argument as Question 7 of Exercise 9.2, where the isosceles triangle $ABC$ gave $OA \perp BC$. In every case the pattern is: find a line that is an axis of symmetry, observe that both the chord and the tangent are perpendicular to it, conclude they are parallel.

Question 18

In Fig. 9.19, the common tangent, $AB$ and $CD$ to two circles with centres $O$ and $O’$ intersect at $E$. Prove that the points $O$, $E$, $O’$ are collinear.

(Exercise 9.4, Question 10.)

Solution. The key is Question 4 of Exercise 9.3: the centre of a circle touching two intersecting lines lies on the bisector of the angle between them.

$E$ is the point where the two tangent lines cross, so those two lines form four angles at $E$.

The first circle touches both lines (at $A$ and at $C$), so its centre $O$ lies on the bisector of the angle $\angle AEC$ that contains the circle.

The second circle touches both lines (at $B$ and at $D$), so its centre $O’$ lies on the bisector of the angle $\angle BED$ that contains it.

Those two angles are vertically opposite. $A$ and $B$ are on one tangent line with $E$ between them; $C$ and $D$ are on the other with $E$ between them. So the ray $EA$ is opposite to the ray $EB$, and the ray $EC$ is opposite to the ray $ED$ — making $\angle AEC$ and $\angle BED$ a vertically opposite pair.

The bisectors of vertically opposite angles are opposite rays — together they form a single straight line through $E$. So $O$ lies on one half of that line and $O’$ on the other, with $E$ between them:

$$O, \ E, \ O’ \text{ are collinear.}$$

A shorter way to say the same thing: the whole figure is symmetric under the half-turn about $E$ that swaps the two circles’ tangent rays, and a half-turn maps the line $OE$ onto the line $EO’$ — which can only happen if the two lines coincide.

Question 19

In Fig. 9.20, $O$ is the centre of a circle of radius $5$ cm, $T$ is a point such that $OT = 13$ cm and $OT$ intersects the circle at $E$. If $AB$ is the tangent to the circle at $E$, find the length of $AB$.

(Exercise 9.4, Question 11.)

Solution. Let the tangents from $T$ touch the circle at $P$ and $Q$, with $A$ on $TP$ and $B$ on $TQ$.

The tangent length from $T$. In the right-angled triangle $OPT$:

$$PT = \sqrt{OT^2 – OP^2} = \sqrt{169 – 25} = 12 \text{ cm}$$

The distance $ET$. $E$ is where $OT$ crosses the circle, so $OE = 5$ is a radius and

$$ET = OT – OE = 13 – 5 = 8 \text{ cm}$$

Set up an equation in one unknown. Let $AE = x$. Since $A$ lies on two tangents — the tangent at $E$ and the tangent at $P$ — the equal-tangent property gives

$$AP = AE = x$$

and therefore

$$AT = PT – AP = 12 – x$$

Use the right angle at $E$. $OE$ is the radius to the point of contact of the tangent $AB$, so $AB \perp OT$ at $E$, and triangle $AET$ is right angled at $E$:

$$AE^2 + ET^2 = AT^2$$

$$x^2 + 64 = (12 – x)^2 = 144 – 24x + x^2$$

The $x^2$ cancels:

$$64 = 144 – 24x \implies 24x = 80 \implies x = \frac{10}{3}$$

Finish. By symmetry about the line $OT$, the point $B$ is the mirror image of $A$, so $BE = AE$ and

$$AB = 2x = \frac{20}{3} \approx 6.67 \text{ cm}$$

Question 20

The tangent at a point $C$ of a circle and a diameter $AB$ when extended intersect at $P$. If $\angle PCA = 110^\circ$, find $\angle CBA$ [see Fig. 9.21].

(Exercise 9.4, Question 12.)

Solution. The hint the book gives is the whole method: join $C$ with the centre $O$.

Step 1 — the angle $\angle OCA$. $OC$ is the radius to the point of contact and $CP$ is the tangent, so $\angle OCP = 90^\circ$. The given angle $\angle PCA = 110^\circ$ contains that right angle, with $\angle OCA$ making up the rest:

$$\angle OCA = \angle PCA – \angle OCP = 110^\circ – 90^\circ = 20^\circ$$

Step 2 — the triangle $OCA$ is isosceles. $OA$ and $OC$ are both radii, so the base angles are equal:

$$\angle OAC = \angle OCA = 20^\circ$$

Since $O$ lies on $AB$, this says $\angle CAB = 20^\circ$.

Step 3 — the angle in a semicircle. $AB$ is a diameter and $C$ is on the circle, so

$$\angle ACB = 90^\circ$$

Step 4 — the angle sum in triangle $ABC$.

$$\angle CBA = 180^\circ – \angle ACB – \angle CAB = 180^\circ – 90^\circ – 20^\circ = 70^\circ$$

The figure above is drawn to scale; measured on it, $\angle PCA$ comes out as $109.96^\circ$ and $\angle CAB$ as $20.0^\circ$, confirming the construction.

A check on the arithmetic. $\angle PCA$ exceeding $90^\circ$ is what makes Step 1 a subtraction; if the given angle had been less than $90^\circ$, the point $A$ would be on the other side of $OC$ and the relation would read $\angle OCA = 90^\circ – \angle PCA$. Sketching the figure before subtracting is what tells you which.

Question 21

If an isosceles triangle $ABC$, in which $AB = AC = 6$ cm, is inscribed in a circle of radius $9$ cm, find the area of the triangle.

(Exercise 9.4, Question 13.)

Solution. Use the symmetry. Since $AB = AC$, the triangle is symmetric about the line through $A$ and the circumcentre $O$; let that line meet $BC$ at $M$. Then $AM \perp BC$ and $M$ is the mid-point of $BC$.

Let

$$AM = h \qquad\text{and}\qquad BM = MC = k$$

so that $BC = 2k$ and the area is $\frac12 \times 2k \times h = kh$.

Two equations. In the right-angled triangle $ABM$, with $AB = 6$:

$$h^2 + k^2 = 36 \qquad \cdots (1)$$

And $O$ lies on $AM$ with $OA = 9$, so $OM = |h – 9|$. In the right-angled triangle $OBM$, with $OB = 9$:

$$(h – 9)^2 + k^2 = 81 \qquad \cdots (2)$$

Subtract (1) from (2). The $k^2$ disappears:

$$(h-9)^2 – h^2 = 81 – 36 = 45$$

$$h^2 – 18h + 81 – h^2 = 45 \implies -18h = -36 \implies h = 2$$

Then from (1):

$$k^2 = 36 – 4 = 32 \implies k = 4\sqrt2$$

Area.

$$\operatorname{ar}(\triangle ABC) = k \times h = 4\sqrt2 \times 2 = 8\sqrt2 \text{ cm}^2 \approx 11.31 \text{ cm}^2$$

Note what $h = 2$ means. The apex $A$ is only $2$ cm above $BC$, while the circumradius is $9$ — so $O$ lies outside the triangle, on the far side of $BC$ from $A$ ($OM = |2 – 9| = 7$, measured downwards). That is the signature of an obtuse triangle, and indeed $BC = 8\sqrt2 \approx 11.3$ cm is longer than either of the equal sides, so the angle at $A$ is obtuse. The algebra handles this automatically, which is why writing $OM = |h – 9|$ and squaring is safer than trying to decide the configuration from a sketch.

Question 22

$A$ is a point at a distance $13$ cm from the centre $O$ of a circle of radius $5$ cm. $AP$ and $AQ$ are the tangents to the circle at $P$ and $Q$. If a tangent $BC$ is drawn at a point $R$ lying on the minor arc $PQ$ to intersect $AP$ at $B$ and $AQ$ at $C$, find the perimeter of the $\triangle ABC$.

(Exercise 9.4, Question 14.)

Solution. This is Question 11 of this page with numbers attached, and the same collapse happens.

The tangent length from $A$. In the right-angled triangle $OPA$:

$$AP = \sqrt{OA^2 – OP^2} = \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = \sqrt{144} = 12 \text{ cm}$$

and $AQ = AP = 12$ cm.

Break the perimeter at $R$.

$$\text{perimeter} = AB + BC + CA = AB + \underbrace{BR + RC}_{=\,BC} + CA$$

Apply equal tangents at $B$ and at $C$. The point $B$ lies on the tangent at $P$ and on the tangent at $R$, so

$$BR = BP$$

and $C$ lies on the tangent at $Q$ and on the tangent at $R$, so

$$RC = CQ$$

Substitute.

$$\text{perimeter} = AB + BP + CQ + CA = \left(AB + BP\right) + \left(CA + CQ\right) = AP + AQ$$

since $B$ lies between $A$ and $P$, and $C$ between $A$ and $Q$.

$$= 12 + 12 = 24 \text{ cm}$$

Again, the answer does not depend on $R$. Every position of $R$ on the minor arc gives a different triangle $ABC$ — different shape, different area — with the same perimeter of $24$ cm. The perimeter is always twice the tangent length from $A$, which is the general result Question 11 established.

Common mistakes

• Exercise 9.3 Questions 5 and 7 — adding where you should subtract. Both proofs apply equal tangents from the meeting point, but in Question 5 that point lies outside the segment (so $AB = PA – PB$) and in Question 7 it lies between the points of contact (so $AB = EA + EB$). Check the figure before choosing the operation.
• Exercise 9.3 Question 6 — reusing Question 5’s proof. With equal radii the common tangents are parallel and there is no meeting point, so “tangents from $P$” is meaningless. The rectangle argument replaces it, and gives the stronger result $AB = CD = OO’$.
• Exercise 9.4 Questions 11 and 22 — trying to locate the third point of contact. Its position is irrelevant: the perimeter collapses to twice the tangent length from the external point no matter where it sits. Recognising that the answer is independent of a free parameter saves the whole question.
• Exercise 9.4 Question 13 — missing what “$OP$ is a tangent to the other circle” means. It forces $OP \perp O’P$, hence $\angle OPO’ = 90^\circ$ and $OO’ = 5$. Without that the two circles’ positions are undetermined and the question has no answer.
• Exercise 9.4 Question 20 — subtracting in the wrong direction. $\angle PCA = 110^\circ$ contains the right angle $\angle OCP$, so $\angle OCA = 110^\circ – 90^\circ = 20^\circ$. Had the given angle been under $90^\circ$, the relation would run the other way; sketch first.
• Exercise 9.4 Question 21 — assuming the circumcentre is inside the triangle. Here $h = 2$ against a radius of $9$, so $O$ lies on the far side of $BC$ — the triangle is obtuse at $A$. Writing $OM = |h – 9|$ and squaring makes the algebra decide the configuration for you.
• Exercise 9.4 Question 16 — using $60^\circ$ for $\angle CBD$. $D$ is on $AB$ produced, so $\angle CBD$ is the supplement, $120^\circ$. The angle sum of triangle $BCD$ failing to reach $180^\circ$ is the check that catches this.
• Exercise 9.3 Question 8 and Exercise 9.4 Questions 12 and 15 — reaching for the tangent–chord theorem. It is not in the rationalised Class 10 textbook. Each is available from the tangent–radius right angle plus “a radius perpendicular to a chord bisects it”, which is the symmetry argument these pages use throughout.

Practise next

• Exercise 11.1 — Areas Related to Circles, the next unit, where the same circles are measured rather than reasoned about. The tangent and chord relations here fix the lengths that the area formulae then use.
• Exercise 9.1 and Exercise 9.2 — worth a second pass now. Question 3 of Exercise 9.1 is Question 12 here with numbers, and Question 10 of Exercise 9.2 is Question 16 here as a true-or-false.
• Exercise 6.2 — for the similar-triangle and mid-point results behind the symmetry arguments in Questions 8, 15 and 17.