NCERT Exemplar Solutions Class 10 Maths Circles — Exercise 9.1

NCERT Exemplar Problems, Class 10 Mathematics — Circles, Exercise 9.1. All 10 multiple-choice questions solved.

The whole of Class 10 Circles rests on two theorems, and this exercise is ten different ways of applying them.

Theorem 1. The tangent at any point of a circle is perpendicular to the radius through the point of contact. So wherever a tangent touches, there is a right angle waiting to be used — and a right angle means Pythagoras.

Theorem 2. The lengths of the two tangents drawn from an external point are equal. So an external point $P$ with tangents to $A$ and $B$ always gives an isosceles triangle $PAB$, and a kite $PAOB$ with right angles at $A$ and $B$.

Between them these two generate everything. In particular, for an external point $P$ at distance $d$ from the centre of a circle of radius $r$, the tangent length is

$$PA = \sqrt{d^2 – r^2}$$

straight from Pythagoras in the right-angled triangle $OAP$.

A note on the current syllabus. The rationalised Class 10 chapter is very short: Introduction, Tangent to a Circle, Number of Tangents from a Point on a Circle, Summary — two theorems and three summary points. In particular the tangent–chord theorem (the “angle in the alternate segment”) is not there; the phrase “alternate segment” occurs nowhere in the chapter.

Questions 3, 7 and 10 here are usually solved with it, and every published solution set does so. This page solves all three by joining the centre instead — which uses only Theorem 1, the isosceles triangle formed by two radii, and the angle sum. That route is longer by two lines and is the one your textbook actually supports.

Key insight. When a question gives you a tangent, draw the radius to the point of contact. When it gives you two tangents from a point, draw both radii. Almost every question in this exercise becomes a right-angled triangle or an isosceles triangle the moment those lines are on the page.

Choose the correct answer from the given four options:

Question 1

If radii of two concentric circles are $4$ cm and $5$ cm, then the length of each chord of one circle which is tangent to the other circle is

Solution. The chord belongs to the bigger circle (radius $5$) and touches the smaller one (radius $4$) — a chord of the small circle could never reach out far enough to touch the big one.

Let the chord be $AB$, touching the small circle at $M$. Since $OM$ is the radius of the small circle drawn to its point of contact, Theorem 1 gives $OM \perp AB$.

And a radius perpendicular to a chord bisects it, so $M$ is the mid-point of $AB$ and $AM = MB$.

Now Pythagoras in the right-angled triangle $OMA$, where $OA = 5$ is the radius of the big circle and $OM = 4$:

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

That is only half the chord. The full chord is

$$AB = 2 \times 3 = 6 \text{ cm}$$

Option (A), $3$ cm, is the half-chord — the single commonest error in this question, and the reason it is worth writing “$AM = 3$, so $AB = 6$” explicitly rather than stopping at the square root.

Question 2

In Fig. 9.3, if $\angle AOB = 125^\circ$, then $\angle COD$ is equal to

Solution. The quadrilateral $ABCD$ has all four sides touching the circle, and $O$ is the centre. The result to use is that in such a figure

$$\angle AOB + \angle COD = 180^\circ$$

so $\angle COD = 180^\circ – 125^\circ = 55^\circ$.

Why that is true is worth a moment, because the result looks arbitrary. Join $O$ to each of the four points where the circle touches a side. Each of the segments $OA$, $OB$, $OC$, $OD$ bisects the angle of the quadrilateral at that vertex, because the two tangents from a vertex are equal in length and $O$ is equidistant from both sides.

That means the four angles at $O$ are

$$\angle AOB = 180^\circ – \frac{\angle A + \angle B}{2}, \qquad \angle COD = 180^\circ – \frac{\angle C + \angle D}{2}$$

each from the angle sum of the triangle formed at that side. Adding them:

$$\angle AOB + \angle COD = 360^\circ – \frac{\angle A + \angle B + \angle C + \angle D}{2} = 360^\circ – \frac{360^\circ}{2} = 180^\circ$$

using the fact that the four angles of a quadrilateral sum to $360^\circ$.

So the two angles subtended at the centre by opposite sides are always supplementary, whatever the shape. Option (A), $62.5^\circ$, is half of $125^\circ$ and is what you get from guessing that the relation is halving rather than supplementing.

Question 3

In Fig. 9.4, $AB$ is a chord of the circle and $AOC$ is its diameter such that $\angle ACB = 50^\circ$. If $AT$ is the tangent to the circle at the point $A$, then $\angle BAT$ is equal to

Solution. Two facts, applied in order.

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

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

In triangle $ABC$ the angles sum to $180^\circ$, so

$$\angle BAC = 180^\circ – 90^\circ – 50^\circ = 40^\circ$$

Second, 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 line $OA$ produced, the angle $\angle CAT$ is that same $90^\circ$, and $\angle BAC$ sits inside it:

$$\angle BAT = \angle CAT – \angle BAC = 90^\circ – 40^\circ = 50^\circ$$

A shortcut you may not use. The tangent–chord theorem says $\angle BAT = \angle ACB$ directly — the angle between a tangent and a chord equals the angle the chord subtends in the alternate segment. That gives $50^\circ$ in one line and explains why the answer equals the given angle exactly. But the theorem is not in the rationalised Class 10 textbook, so the two-step route above is the one to write out.

Option (D), $40^\circ$, is $\angle BAC$ — the angle you find on the way, not the one asked for.

Question 4

From a point $P$ which is at a distance of $13$ cm from the centre $O$ of a circle of radius $5$ cm, the pair of tangents $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is

Solution. First the tangent length. In the right-angled triangle $OQP$ — right angled at $Q$, because $OQ$ is the radius to the point of contact:

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

the $5$–$12$–$13$ triangle.

The quadrilateral $PQOR$ is made of two triangles, $\triangle OQP$ and $\triangle ORP$, joined along $OP$. They are congruent — $OQ = OR = 5$, $PQ = PR = 12$, and $OP$ is common — so each has area

$$\frac12 \times PQ \times OQ = \frac12 \times 12 \times 5 = 30 \text{ cm}^2$$

taking the two legs of the right angle as base and height. Together:

$$\operatorname{ar}(PQOR) = 2 \times 30 = 60 \text{ cm}^2$$

Option (C), $30$, is one triangle — the same halving error as Question 1, and worth naming because it is the one this exercise keeps setting.

Question 5

At one end $A$ of a diameter $AB$ of a circle of radius $5$ cm, tangent $XAY$ is drawn to the circle. The length of the chord $CD$ parallel to $XY$ and at a distance $8$ cm from $A$ is

Solution. The distance in this question is measured from $A$, not from the centre, and converting between the two is the whole problem.

The tangent $XY$ at $A$ is perpendicular to the diameter $AB$. The chord $CD$ is parallel to $XY$, so $CD$ is also perpendicular to $AB$, and its distance from $A$ is measured along $AB$.

The centre $O$ lies on $AB$, at $5$ cm from $A$ (one radius). The chord is $8$ cm from $A$ along the same line. So the distance from the centre to the chord is

$$8 – 5 = 3 \text{ cm}$$

Now drop the perpendicular $OM$ from the centre to the chord; it bisects $CD$. By Pythagoras in the right-angled triangle $OMC$, with $OC = 5$ the radius:

$$CM = \sqrt{5^2 – 3^2} = \sqrt{25 – 9} = 4$$

and the full chord is

$$CD = 2 \times 4 = 8 \text{ cm}$$

Two checks that the set-up is right. The chord is $8$ cm from $A$ and the diameter is $10$ cm, so the chord lies between $A$ and $B$ — it exists. And $3 < 5$, so the perpendicular distance from the centre is less than the radius, which it must be for a chord.

Option (A), $4$ cm, is the half-chord again.

Question 6

In Fig. 9.5, $AT$ is a tangent to the circle with centre $O$ such that $OT = 4$ cm and $\angle OTA = 30^\circ$. Then $AT$ is equal to

Solution. $OA$ is the radius to the point of contact $A$, so $OA \perp AT$ and triangle $OAT$ is right angled at $A$. That makes $OT$ the hypotenuse.

Relative to the given angle of $30^\circ$ at $T$, the side $AT$ is adjacent and $OT$ is the hypotenuse, so the ratio to use is the cosine:

$$\cos 30^\circ = \frac{AT}{OT} \implies AT = 4\cos 30^\circ = 4 \times \frac{\sqrt3}{2} = 2\sqrt3 \text{ cm}$$

Option (B), $2$ cm, is $OA = 4\sin 30^\circ$ — the radius, which the figure does not ask for. Getting it means putting the $30^\circ$ at the wrong vertex or reaching for the sine instead of the cosine.

A check: $2\sqrt3 \approx 3.46$, which is less than the hypotenuse $4$ and greater than the other leg $2$, as a leg opposite the larger of the two acute angles must be.

Question 7

In Fig. 9.6, if $O$ is the centre of a circle, $PQ$ is a chord and the tangent $PR$ at $P$ makes an angle of $50^\circ$ with $PQ$, then $\angle POQ$ is equal to

Solution. Join $O$ to $P$ and to $Q$ — the standard move, and the one that makes this question routine.

At $P$. $OP$ is the radius to the point of contact and $PR$ is the tangent, so $\angle OPR = 90^\circ$. The chord $PQ$ sits inside that right angle, making $50^\circ$ with $PR$, so

$$\angle OPQ = 90^\circ – 50^\circ = 40^\circ$$

In triangle $OPQ$. Both $OP$ and $OQ$ are radii, so the triangle is isosceles and the base angles are equal:

$$\angle OQP = \angle OPQ = 40^\circ$$

Angle sum.

$$\angle POQ = 180^\circ – 40^\circ – 40^\circ = 100^\circ$$

Option (B), $80^\circ$, is $\angle OPQ + \angle OQP$ — the two base angles added rather than subtracted from $180^\circ$.

The tangent–chord theorem would give this faster: the $50^\circ$ equals the inscribed angle in the alternate segment, and the central angle is twice any inscribed angle on the same arc, so $\angle POQ = 100^\circ$. That theorem is not in the current Class 10 textbook, and the isosceles-triangle route above needs nothing beyond it.

Question 8

In Fig. 9.7, if $PA$ and $PB$ are tangents to the circle with centre $O$ such that $\angle APB = 50^\circ$, then $\angle OAB$ is equal to

Solution. Two steps, each using one of the two theorems.

Theorem 2 gives the triangle $PAB$. The tangents from $P$ are equal, so $PA = PB$ and triangle $PAB$ is isosceles. Its base angles are therefore equal, and the angle sum gives

$$\angle PAB = \angle PBA = \frac{180^\circ – 50^\circ}{2} = \frac{130^\circ}{2} = 65^\circ$$

Theorem 1 gives the right angle at $A$. $OA$ is the radius to the point of contact and $PA$ is the tangent, so

$$\angle OAP = 90^\circ$$

The angle asked for sits inside that right angle, with $\angle PAB$ taking up part of it:

$$\angle OAB = \angle OAP – \angle PAB = 90^\circ – 65^\circ = 25^\circ$$

Notice the relationship that falls out: $\angle OAB = 25^\circ$ is exactly half of $\angle APB = 50^\circ$. That is true in general — $\angle OAB = \frac12\angle APB$ for any pair of tangents — and it is worth carrying, because this configuration recurs throughout the unit.

Option (C), $40^\circ$, comes from taking $\angle PAB$ as $50^\circ$; option (D) is the given angle repeated.

Question 9

If two tangents inclined at an angle $60^\circ$ are drawn to a circle of radius $3$ cm, then length of each tangent is equal to

Solution. Let the tangents run from the external point $P$ to the points of contact $A$ and $B$, with $\angle APB = 60^\circ$.

The line $OP$ bisects that angle. Triangles $OAP$ and $OBP$ are congruent — $OA = OB$ (radii), $PA = PB$ (Theorem 2), $OP$ common — so

$$\angle APO = \frac{60^\circ}{2} = 30^\circ$$

Now work in the right-angled triangle $OAP$, which is right angled at $A$ by Theorem 1. Relative to the $30^\circ$ angle at $P$, the radius $OA = 3$ is opposite and the tangent $PA$ is adjacent, so the ratio to use is the tangent:

$$\tan 30^\circ = \frac{OA}{PA} \implies \frac{1}{\sqrt3} = \frac{3}{PA} \implies PA = 3\sqrt3 \text{ cm}$$

$3\sqrt3 \approx 5.20$ cm.

Halving the angle is the step to get right. Using $60^\circ$ instead of $30^\circ$ gives $PA = \frac{3}{\sqrt3} = \sqrt3 \approx 1.73$ cm, which is shorter than the radius — and a tangent from an external point is always longer than that, since $OP > r$ forces $PA = \sqrt{OP^2 – r^2}$ to grow without bound as $P$ moves out. A tangent shorter than the radius should make you re-check.

Option (B), $6$ cm, is $OP$: the distance to the centre, since $OP = \frac{3}{\sin 30^\circ} = 6$.

Question 10

In Fig. 9.8, if $PQR$ is the tangent to a circle at $Q$ whose centre is $O$, $AB$ is a chord parallel to $PR$ and $\angle BQR = 70^\circ$, then $\angle AQB$ is equal to

Solution. The parallel chord is what makes this work, so use it first.

Step 1 — transfer the angle across the parallels. $AB \parallel PR$, and $QB$ is a transversal cutting both. So $\angle ABQ$ and $\angle BQR$ are alternate angles:

$$\angle ABQ = \angle BQR = 70^\circ$$

Step 2 — the triangle $AQB$ is isosceles. Here is why, without the tangent–chord theorem. Draw $OQ$; it is perpendicular to the tangent $PR$, hence perpendicular to $AB$ as well, since $AB \parallel PR$. A radius perpendicular to the chord $AB$ bisects it, so $OQ$ passes through the mid-point of $AB$ and the whole figure is symmetric about the line $OQ$.

That symmetry swaps $A$ with $B$ and fixes $Q$, so

$$QA = QB \quad\text{and}\quad \angle QAB = \angle QBA = 70^\circ$$

Step 3 — angle sum in triangle $AQB$.

$$\angle AQB = 180^\circ – 70^\circ – 70^\circ = 40^\circ$$

Option (A), $20^\circ$, is half the answer; option (C), $35^\circ$, is half the given angle. Both come from guessing at a halving rather than following the triangle through.

Published solutions do Step 2 with the tangent–chord theorem: $\angle BQR = \angle QAB$ because they stand on the same chord $QB$ in alternate segments. The symmetry argument above reaches the same place using only the perpendicular from the centre to a chord.

Common mistakes

• Questions 1, 4 and 5 — stopping at half. Pythagoras in the triangle formed by the centre, the mid-point of a chord and one end gives the half-chord, and a tangent pair gives one of two congruent triangles. Options (A) in Question 1, (C) in Question 4 and (A) in Question 5 are all exactly half the answer.
• Question 5 — measuring the chord’s distance from the wrong point. The $8$ cm is from $A$, on the circle; the perpendicular distance from the centre is $8 – 5 = 3$ cm, and that is the number Pythagoras needs.
• Question 9 — using the full angle between the tangents. The line from the external point to the centre bisects that angle, so the right-angled triangle contains $30^\circ$, not $60^\circ$. A tangent length shorter than the radius is the symptom.
• Questions 3, 6 and 7 — losing track of which angle is the right angle. The tangent is perpendicular to the radius at the point of contact, so in $\triangle OAT$ the right angle is at $A$ and $OT$ is the hypotenuse. Putting it at $T$ or at $O$ makes every ratio wrong.
• Questions 3 and 8 — answering the intermediate angle. In Question 3 the working produces $\angle BAC = 40^\circ$ before the answer $50^\circ$, and in Question 8 it produces $\angle PAB = 65^\circ$ before the answer $25^\circ$. Both intermediates appear among the options in similar questions; read what was asked before choosing.
• Question 2 — halving instead of supplementing. The two angles subtended at the centre by opposite sides of a tangential quadrilateral add to $180^\circ$, so $\angle COD = 55^\circ$. Option (A), $62.5^\circ$, is half of $125^\circ$ and tempts anyone reaching for the nearest plausible operation.
• Questions 3, 7 and 10 — relying on the tangent–chord theorem. It is not in the rationalised Class 10 textbook. All three can be done by joining the centre and using the isosceles triangle formed by two radii, which is two lines longer and entirely on-syllabus.

Practise next

• Exercise 9.2 — ten true-or-false statements on the same two theorems, several of which turn on the tangent length $\sqrt{d^2 – r^2}$ and what happens to it as the external point moves.
• Exercises 9.3 and 9.4 — twenty-two questions where these results are proved rather than applied, including the tangent-length identities for circumscribed polygons.
• Exercise 8.1 — worth revisiting if Questions 6 and 9 were slow. Every tangent-and-radius configuration is a right-angled triangle, and choosing between sine, cosine and tangent is the same skill there.