NCERT Class 10 Mathematics — Circles, Exercise 10.2. All 13 questions solved.

Exercise 10.2 is where the chapter’s two theorems start doing real work. Three questions are multiple choice, three ask for a length, and seven are proofs, but all thirteen run on the same two facts:

• Theorem 10.1. The tangent at any point of a circle is perpendicular to the radius through the point of contact.
• Theorem 10.2. The lengths of the two tangents drawn from an external point to a circle are equal.

A consequence of the second is used almost as often as the theorems themselves. If $\mathrm{PA}$ and $\mathrm{PB}$ are the tangents from $\mathrm{P}$ to a circle with centre $\mathrm{O}$, then triangles $\mathrm{OAP}$ and $\mathrm{OBP}$ are congruent (right angles at $\mathrm{A}$ and $\mathrm{B}$, $\mathrm{OA} = \mathrm{OB}$, $\mathrm{OP}$ common). So **$\mathrm{OP}$ bisects both $\angle \mathrm{APB}$ and $\angle \mathrm{AOB}$**.

Key insight. Two moves solve this exercise. Whenever a tangent appears, join the centre to its point of contact: that creates a right angle. Whenever two tangents meet at a point, mark the two tangent lengths from that point as equal. Questions 1 to 7 need only the first move; questions 8, 11 and 12 need only the second; questions 9 and 13 need both.

In Q.1 to 3, choose the correct option and give justification.

Question 1

From a point $\mathrm{Q}$, the length of the tangent to a circle is $24$ cm and the distance of $\mathrm{Q}$ from the centre is $25$ cm. The radius of the circle is

Solution. Let the tangent from $\mathrm{Q}$ touch the circle at $\mathrm{P}$, and let $\mathrm{O}$ be the centre. The radius $\mathrm{OP}$ is perpendicular to the tangent $\mathrm{PQ}$ (Theorem 10.1), so triangle $\mathrm{OPQ}$ is right-angled at $\mathrm{P}$ with hypotenuse $\mathrm{OQ} = 25$ cm:

$$\mathrm{OP}^2 = \mathrm{OQ}^2 – \mathrm{PQ}^2 = 25^2 – 24^2 = (25 – 24)(25 + 24) = 49$$

$$\mathrm{OP} = 7\ \text{cm}$$

Question 2

In Fig. 10.11, if $\mathrm{TP}$ and $\mathrm{TQ}$ are the two tangents to a circle with centre $\mathrm{O}$ so that $\angle \mathrm{POQ} = 110^\circ$, then $\angle \mathrm{PTQ}$ is equal to

Solution. $\mathrm{OP}$ and $\mathrm{OQ}$ are radii to the points of contact, so $\angle \mathrm{OPT} = \angle \mathrm{OQT} = 90^\circ$ (Theorem 10.1). The four angles of quadrilateral $\mathrm{OPTQ}$ add up to $360^\circ$:

$$\angle \mathrm{PTQ} = 360^\circ – 90^\circ – 90^\circ – 110^\circ = 70^\circ$$

The two right angles always use up $180^\circ$, so the angle at $\mathrm{T}$ and the angle at $\mathrm{O}$ always add to $180^\circ$. Question 10 asks you to prove exactly this.

Question 3

If tangents $\mathrm{PA}$ and $\mathrm{PB}$ from a point $\mathrm{P}$ to a circle with centre $\mathrm{O}$ are inclined to each other at angle of $80^\circ$, then $\angle \mathrm{POA}$ is equal to

Solution. The configuration is the same as Fig. 10.11, with $\mathrm{P}$ now the external point and $\mathrm{A}$, $\mathrm{B}$ the points of contact. We are given $\angle \mathrm{APB} = 80^\circ$.

$\mathrm{OP}$ bisects the angle between the tangents, because triangles $\mathrm{OAP}$ and $\mathrm{OBP}$ are congruent by RHS: $\angle \mathrm{OAP} = \angle \mathrm{OBP} = 90^\circ$, $\mathrm{OA} = \mathrm{OB}$ (radii) and $\mathrm{OP}$ is common. Hence

$$\angle \mathrm{APO} = \tfrac{1}{2} \times 80^\circ = 40^\circ$$

In triangle $\mathrm{OAP}$ the angle at $\mathrm{A}$ is $90^\circ$, so

$$\angle \mathrm{POA} = 180^\circ – 90^\circ – 40^\circ = 50^\circ$$

Question 4

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

Solution. Let $\mathrm{AB}$ be a diameter of a circle with centre $\mathrm{O}$, and let $l$ and $m$ be the tangents at $\mathrm{A}$ and $\mathrm{B}$.

By Theorem 10.1, $\mathrm{OA} \perp l$ and $\mathrm{OB} \perp m$. Since $\mathrm{AB}$ is a diameter, $\mathrm{O}$ lies on $\mathrm{AB}$, so $\mathrm{OA}$ and $\mathrm{OB}$ are both parts of the one line $\mathrm{AB}$. Hence the line $\mathrm{AB}$ is perpendicular to both $l$ and $m$.

Now $\mathrm{AB}$ is a transversal to $l$ and $m$, and the two interior angles it makes on the same side are $90^\circ + 90^\circ = 180^\circ$. Interior angles on the same side of a transversal adding up to $180^\circ$ is exactly the condition for two lines to be parallel.

Question 5

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

Solution. Let $\mathrm{XY}$ be the tangent at $\mathrm{P}$ to a circle with centre $\mathrm{O}$.

By Theorem 10.1, the radius $\mathrm{OP}$ is perpendicular to $\mathrm{XY}$. But through a given point $\mathrm{P}$ of the line $\mathrm{XY}$ there is only one line perpendicular to $\mathrm{XY}$. So the perpendicular to $\mathrm{XY}$ at $\mathrm{P}$ must be the line $\mathrm{OP}$ itself, and that line passes through $\mathrm{O}$.

Put another way: if the perpendicular at $\mathrm{P}$ were some other line $\mathrm{PR}$ not through $\mathrm{O}$, then $\angle \mathrm{OPY}$ and $\angle \mathrm{RPY}$ would both be $90^\circ$ with $\mathrm{O}$ and $\mathrm{R}$ on the same side of $\mathrm{XY}$, which is only possible if $\mathrm{PR}$ and $\mathrm{PO}$ are the same ray.

Question 6

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

Solution. Let the tangent from $\mathrm{A}$ touch the circle at $\mathrm{T}$. Then $\mathrm{OT} \perp \mathrm{AT}$, so triangle $\mathrm{OTA}$ is right-angled at $\mathrm{T}$ with hypotenuse $\mathrm{OA} = 5$ cm:

$$\mathrm{OT}^2 = \mathrm{OA}^2 – \mathrm{AT}^2 = 25 – 16 = 9, \qquad \mathrm{OT} = 3\ \text{cm}$$

This is question 1 again with smaller numbers — the $3, 4, 5$ triangle in place of $7, 24, 25$.

Question 7

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.

Solution. Let the chord $\mathrm{AB}$ of the larger circle touch the smaller circle at $\mathrm{P}$, and let $\mathrm{O}$ be the common centre.

$\mathrm{AB}$ is a tangent to the smaller circle at $\mathrm{P}$, so $\mathrm{OP} \perp \mathrm{AB}$ and $\mathrm{OP} = 3$ cm. But $\mathrm{AB}$ is also a chord of the larger circle, and the perpendicular from the centre to a chord bisects the chord. So $\mathrm{P}$ is the midpoint of $\mathrm{AB}$.

In right triangle $\mathrm{OPB}$, with $\mathrm{OB} = 5$ cm the radius of the larger circle:

$$\mathrm{PB} = \sqrt{\mathrm{OB}^2 – \mathrm{OP}^2} = \sqrt{25 – 9} = 4\ \text{cm}$$

$$\mathrm{AB} = 2\,\mathrm{PB} = 8\ \text{cm}$$

Question 8

A quadrilateral $\mathrm{ABCD}$ is drawn to circumscribe a circle (see Fig. 10.12). Prove that

$$\mathrm{AB} + \mathrm{CD} = \mathrm{AD} + \mathrm{BC}$$

Solution. The circle touches $\mathrm{AB}$, $\mathrm{BC}$, $\mathrm{CD}$ and $\mathrm{DA}$ at $\mathrm{P}$, $\mathrm{Q}$, $\mathrm{R}$ and $\mathrm{S}$. Each vertex is an external point with two tangents drawn from it, and by Theorem 10.2 those two tangents are equal:

$$\mathrm{AP} = \mathrm{AS}, \qquad \mathrm{BP} = \mathrm{BQ}, \qquad \mathrm{CR} = \mathrm{CQ}, \qquad \mathrm{DR} = \mathrm{DS}$$

Adding the four equations:

$$\mathrm{AP} + \mathrm{BP} + \mathrm{CR} + \mathrm{DR} = \mathrm{AS} + \mathrm{BQ} + \mathrm{CQ} + \mathrm{DS}$$

Now regroup each side into whole sides of the quadrilateral. On the left, $\mathrm{AP} + \mathrm{BP} = \mathrm{AB}$ and $\mathrm{CR} + \mathrm{DR} = \mathrm{CD}$. On the right, $\mathrm{AS} + \mathrm{DS} = \mathrm{AD}$ and $\mathrm{BQ} + \mathrm{CQ} = \mathrm{BC}$. Therefore

$$\mathrm{AB} + \mathrm{CD} = \mathrm{AD} + \mathrm{BC}$$

Each of the eight tangent segments is counted once on each side, which is why the two sums of opposite sides must agree.

Question 9

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

Solution. Let $\mathrm{XY}$ touch the circle at $\mathrm{P}$ and $\mathrm{X’Y’}$ at $\mathrm{Q}$, and join $\mathrm{OC}$ (dashed in the figure).

Step 1 — $\mathrm{PQ}$ is a diameter. $\mathrm{OP} \perp \mathrm{XY}$ and $\mathrm{OQ} \perp \mathrm{X’Y’}$ by Theorem 10.1. Since $\mathrm{XY} \parallel \mathrm{X’Y’}$, both $\mathrm{OP}$ and $\mathrm{OQ}$ are perpendicular to the same direction and both pass through $\mathrm{O}$, so $\mathrm{P}$, $\mathrm{O}$, $\mathrm{Q}$ lie on one straight line. Hence $\angle \mathrm{POC} + \angle \mathrm{COQ} = 180^\circ$.

Step 2 — $\mathrm{OA}$ and $\mathrm{OB}$ are bisectors. In triangles $\mathrm{OPA}$ and $\mathrm{OCA}$: $\mathrm{OP} = \mathrm{OC}$ (radii), $\mathrm{AP} = \mathrm{AC}$ (tangents from $\mathrm{A}$, Theorem 10.2), and $\mathrm{OA}$ is common. So the triangles are congruent by SSS, and

$$\angle \mathrm{POA} = \angle \mathrm{AOC}$$

In exactly the same way, triangles $\mathrm{OQB}$ and $\mathrm{OCB}$ are congruent, so $\angle \mathrm{QOB} = \angle \mathrm{BOC}$.

Step 3 — add up. The straight angle $\mathrm{POQ}$ is made of four pieces:

$$\angle \mathrm{POA} + \angle \mathrm{AOC} + \angle \mathrm{COB} + \angle \mathrm{BOQ} = 180^\circ$$

$$2\,\angle \mathrm{AOC} + 2\,\angle \mathrm{COB} = 180^\circ$$

$$\angle \mathrm{AOC} + \angle \mathrm{COB} = 90^\circ$$

Since $\mathrm{C}$ lies on the segment $\mathrm{AB}$, $\angle \mathrm{AOC} + \angle \mathrm{COB} = \angle \mathrm{AOB}$, and so $\angle \mathrm{AOB} = 90^\circ$.

Question 10

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.

Solution. Let $\mathrm{PA}$ and $\mathrm{PB}$ be the tangents from an external point $\mathrm{P}$ to a circle with centre $\mathrm{O}$ — the picture is Fig. 10.11 with $\mathrm{T}$ renamed $\mathrm{P}$. We must show $\angle \mathrm{APB} + \angle \mathrm{AOB} = 180^\circ$.

By Theorem 10.1, $\angle \mathrm{OAP} = 90^\circ$ and $\angle \mathrm{OBP} = 90^\circ$. The angles of quadrilateral $\mathrm{OAPB}$ add up to $360^\circ$, so

$$\angle \mathrm{APB} + \angle \mathrm{AOB} = 360^\circ – 90^\circ – 90^\circ = 180^\circ$$

Question 2 was a numerical instance of this: $110^\circ + 70^\circ = 180^\circ$.

Question 11

Prove that the parallelogram circumscribing a circle is a rhombus.

Solution. Let $\mathrm{ABCD}$ be a parallelogram circumscribing a circle. It is in particular a quadrilateral circumscribing a circle, so the result of question 8 applies (the same four pairs of equal tangents give it):

$$\mathrm{AB} + \mathrm{CD} = \mathrm{AD} + \mathrm{BC}$$

In a parallelogram, opposite sides are equal: $\mathrm{CD} = \mathrm{AB}$ and $\mathrm{AD} = \mathrm{BC}$. Substituting,

$$2\,\mathrm{AB} = 2\,\mathrm{BC} \quad\Longrightarrow\quad \mathrm{AB} = \mathrm{BC}$$

So two adjacent sides are equal, and with opposite sides already equal, all four sides are equal: $\mathrm{AB} = \mathrm{BC} = \mathrm{CD} = \mathrm{DA}$. A parallelogram with all sides equal is a rhombus.

Question 12

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

Solution. Let the circle touch $\mathrm{CA}$ at $\mathrm{E}$ and $\mathrm{AB}$ at $\mathrm{F}$. Tangents from the same point are equal (Theorem 10.2):

$$\mathrm{CE} = \mathrm{CD} = 6, \qquad \mathrm{BF} = \mathrm{BD} = 8, \qquad \mathrm{AE} = \mathrm{AF} = x \ \text{(say)}$$

So the sides, in cm, are

$$\mathrm{BC} = 14, \qquad \mathrm{AB} = x + 8, \qquad \mathrm{AC} = x + 6$$

The only unknown is $x$, and the radius has not been used yet. It enters through the area, which we can find in two ways.

Area from the radius. Join $\mathrm{O}$ to $\mathrm{A}$, $\mathrm{B}$, $\mathrm{C}$. This splits the triangle into triangles $\mathrm{OBC}$, $\mathrm{OCA}$ and $\mathrm{OAB}$, and each has height $4$ (the radius to a point of contact is perpendicular to that side). The semi-perimeter is $s = \tfrac{1}{2}(14 + x + 8 + x + 6) = 14 + x$, so

$$\text{ar}(\mathrm{ABC}) = \tfrac{1}{2} \times 4 \times (\mathrm{BC} + \mathrm{CA} + \mathrm{AB}) = 4s = 4(14 + x)$$

Area from Heron’s formula. With $s – \mathrm{BC} = x$, $s – \mathrm{CA} = 8$ and $s – \mathrm{AB} = 6$:

$$\text{ar}(\mathrm{ABC}) = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{(14 + x)\cdot x \cdot 8 \cdot 6} = \sqrt{48x(14 + x)}$$

Equate and solve.

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

$$48x(14 + x) = 16(14 + x)^2$$

Divide by $16(14 + x)$, which is positive:

$$3x = 14 + x \quad\Longrightarrow\quad x = 7$$

Hence $\mathrm{AB} = 7 + 8 = 15$ cm and $\mathrm{AC} = 7 + 6 = 13$ cm.

Check. A $13, 14, 15$ triangle has $s = 21$ and area $\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$, and $84 = 4 \times 21$, so its inscribed circle does have radius $4$.

Question 13

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

Solution. Let $\mathrm{ABCD}$ circumscribe a circle with centre $\mathrm{O}$, touching $\mathrm{AB}$, $\mathrm{BC}$, $\mathrm{CD}$, $\mathrm{DA}$ at $\mathrm{P}$, $\mathrm{Q}$, $\mathrm{R}$, $\mathrm{S}$. Join $\mathrm{O}$ to the four vertices and to the four points of contact. We must show $\angle \mathrm{AOB} + \angle \mathrm{COD} = 180^\circ$ and $\angle \mathrm{BOC} + \angle \mathrm{AOD} = 180^\circ$.

Step 1 — four pairs of equal angles. In triangles $\mathrm{OAP}$ and $\mathrm{OAS}$: $\mathrm{OP} = \mathrm{OS}$ (radii), $\mathrm{AP} = \mathrm{AS}$ (tangents from $\mathrm{A}$) and $\mathrm{OA}$ is common, so they are congruent (SSS) and $\angle \mathrm{AOP} = \angle \mathrm{AOS}$. The same argument at the other three vertices gives

$$\angle \mathrm{BOP} = \angle \mathrm{BOQ}, \qquad \angle \mathrm{COQ} = \angle \mathrm{COR}, \qquad \angle \mathrm{DOR} = \angle \mathrm{DOS}$$

Call the four common values $a$, $b$, $c$ and $d$ respectively (with $a = \angle \mathrm{AOP} = \angle \mathrm{AOS}$).

Step 2 — the full turn at O. The eight angles round $\mathrm{O}$ make up $360^\circ$:

$$2a + 2b + 2c + 2d = 360^\circ \quad\Longrightarrow\quad a + b + c + d = 180^\circ$$

Step 3 — read off the opposite sides. Side $\mathrm{AB}$ subtends $\angle \mathrm{AOB} = a + b$ and side $\mathrm{CD}$ subtends $\angle \mathrm{COD} = c + d$. So

$$\angle \mathrm{AOB} + \angle \mathrm{COD} = a + b + c + d = 180^\circ$$

Likewise $\angle \mathrm{BOC} = b + c$ and $\angle \mathrm{AOD} = a + d$, which also add to $180^\circ$.

Common mistakes

• Question 1, putting the right angle at the centre. The right angle is between the radius and the tangent, at the point of contact. Treat $25$ as a leg instead of the hypotenuse and you get $\sqrt{25^2 + 24^2} \approx 34.7$, which is not even an option.
• Question 3, stopping at $40^\circ$. $40^\circ$ is $\angle \mathrm{APO}$, half the angle between the tangents. The question asks for the angle at the centre, $\angle \mathrm{POA}$, which is its complement in the right triangle.
• Question 7, giving $4$ cm. That is only half the chord. The perpendicular from the centre bisects the chord, so the whole chord is twice $\mathrm{PB}$.
• Question 8, pairing the wrong segments. Theorem 10.2 pairs the two tangents from the same vertex — $\mathrm{AP}$ with $\mathrm{AS}$, not $\mathrm{AP}$ with $\mathrm{BP}$. Pairing the two halves of one side proves nothing.
• Question 9, skipping the diameter. The proof needs $\mathrm{P}$, $\mathrm{O}$ and $\mathrm{Q}$ to lie on a straight line, and that follows from the tangents being parallel. Students who use the $180^\circ$ without saying why have left out the one step that uses the hypothesis $\mathrm{XY} \parallel \mathrm{X’Y’}$.
• Question 12, assuming the triangle is isosceles or right-angled. Nothing in the question says so. The unknown tangent length $x$ from $\mathrm{A}$ is what makes the problem solvable, and the radius is used only through the area.

Practise next

• Chapter 11, Exercise 11.1 — sectors and segments, where the triangle made by two radii and a chord from question 7 is needed in almost every question.
• Chapter 6, Exercise 6.3 — AA similarity, which the chapter’s own worked example uses as an alternative to Pythagoras for finding a tangent length.
• Exercise 10.1 — revise the definitions and the tangent–radius right angle that every question here depends on.