Tag: Exercise 10.2 Chapter 10 NCERT Answers and Solutions

Q1

Question. 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

Solution.
Let centre be $O$, point of tangency be $T$. In right triangle $OTQ$ (radius $OT$ ⟂ tangent at $T$):

$$O{Q}^2=O{T}^2+T{Q}^2\mathrm{}$$

So ${25}^2=r^2+{24}^2$. Hence

$$r^2=625-576=49\Rightarrow r=7\text{ cm}\mathrm{.}$$

Answer: (A) 7 cm.

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Q2

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

Solution.
The angle between the two tangents = ${180}^{\circ }$ (central angle between their points of contact). So

$$\mathrm{\angle }PTQ={180}^{\circ }-\mathrm{\angle }POQ={180}^{\circ }-{110}^{\circ }={70}^{\circ }\mathrm{.}$$

Answer: (B) ${70}^{\circ }$.

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Q3

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

Solution.
Let $\mathrm{\angle }APB={80}^{\circ }$. The radius $OP$ bisects the angle between the tangents, so $\mathrm{\angle }OPA={40}^{\circ }$. In right triangle $OAP$, angle at $A$ is ${90}^{\circ }$ (radius ⟂ tangent), so the three angles are ${90}^{\circ },{40}^{\circ },\mathrm{\angle }POA$. Thus

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

Answer: (A) ${50}^{\circ }$.

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Q4

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

Solution.
Let diameter be $AB$ and tangents at $A$ and $B$ be $t_A$ and $t_B$. Radius $OA$ is collinear with $OB$ (they form the diameter). Tangent at $A$ is perpendicular to $OA$; tangent at $B$ is perpendicular to $OB$. Since $OA$ and $OB$ lie on the same straight line, the two perpendiculars are parallel. Hence tangents at the ends of a diameter are parallel.

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Q5

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

Solution.
Let circle centre $O$ and tangent at $P$ be line $t$. By Theorem 10.1, the radius $OP$ is perpendicular to the tangent at the point of contact $P$. Hence the perpendicular to the tangent at $P$ is exactly the radius $OP$, so it passes through the centre. ∎

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Q6

Question. 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.

Solution.
Let radius $=r$. In right triangle $OAT$:

$$O{A}^2=O{T}^2+A{T}^2\Rightarrow 5^2=r^2+4^2\mathrm{.}$$

So $r^2=25-16=9\Rightarrow r=3$

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Q7

Question. 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 of larger circle touch the smaller at $P$. Distance from centre to the chord = radius of smaller circle $=3$. Half-chord length $=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4$. So full chord length $=2\times 4=8$ cm.

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Q8

Question. A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig.10.12). Prove that

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

Solution.
Let the circle touch $AB,BC,CD,DA$ at $P,Q,R,S$ respectively. Tangent segments from a vertex to the circle are equal, so

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

Compute:

$$AB+CD=(AP+BP)+(CR+DR)=(AS+BQ)+(CQ+DS)\mathrm{.}$$

Using equalities $BP=BQ,CR=CQ,AP=AS,DR=DS$ we rearrange to get

$$AB+CD=(AS+DS)+(BQ+CQ)=AD+BC\mathrm{.}$$

Thus $AB+CD=AD+BC$

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Q9

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

Solution. (coordinate / analytic proof — short and clean)
Place the circle with centre $O(0,0)$ and radius $r$. Take the two parallel tangents as the horizontal lines $y=r$ and $y=-r$. Then the tangent at a point $C(\cos \theta ,\sin \theta )$ has equation

$$x\cos \theta +y\sin \theta =r\mathrm{.}$$

Its intersection $A$ with $y=r$ is at

$$x_A=\frac{r-r\sin \theta }{\cos \theta }=\frac{r(1-\sin \theta )}{\cos \theta },A=(\frac{r(1-\sin \theta )}{\cos \theta },r)\mathrm{.}$$

Its intersection $B$ with $y=-r$ is

$$x_B=\frac{r+r\sin \theta }{\cos \theta }=\frac{r(1+\sin \theta )}{\cos \theta },B=(\frac{r(1+\sin \theta )}{\cos \theta },-r)\mathrm{.}$$

Compute dot product of vectors $\overrightarrow{OA}$and $\overrightarrow{OB}$:

$$\overrightarrow{OA}\cdot \overrightarrow{OB}=\frac{r^2(1-{\sin }^2\theta )}{{\cos }^2\theta }+(r)(-r)=\frac{r^2{\cos }^2\theta }{{\cos }^2\theta }-r^2=r^2-r^2=0.$$

So $OA\perp OB$, therefore $\mathrm{\angle }AOB={90}^{\circ }$.

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Q10

Question. 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 the tangents from external point $P$ touch the circle at $A$ and $B$. Then (standard result)

$$\mathrm{\angle }APB={180}^{\circ }-\mathrm{\angle }AOB,$$

because each tangent is perpendicular to the radius at the point of contact; triangle $OAP$ and $OBP$ give $\mathrm{\angle }OPA=\frac{1}{2}\mathrm{\angle }APB$ and $\mathrm{\angle }OPA+\mathrm{\angle }POA+{90}^{\circ }={180}^{\circ }$ etc. So $\mathrm{\angle }APB+\mathrm{\angle }AOB={180}^{\circ }$. Hence they are supplementary. ∎

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Q11

Question. Prove that the parallelogram circumscribing a circle is a rhombus.

Solution.
If a parallelogram circumscribes a circle, then by Q8 we have for the circumscribed quadrilateral $ABCD$:

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

But in a parallelogram $AB=CD$ and $BC=AD$. Substitute to get:

$$AB+AB=AD+AD\Rightarrow 2AB=2AD\Rightarrow AB=AD\mathrm{.}$$

Thus two adjacent sides are equal; in a parallelogram equal adjacent sides imply all four sides are equal. Therefore the parallelogram is a rhombus. ∎

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Q12

Question. 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 $AC$.

Solution.
Let the circle touch $AB,BC,CA$ at $F,D,E$ respectively. Tangent segments from same vertex are equal: let $AF=AE=x$. From given,

$$BF=BD=8,CE=CD=6.$$

Thus

$$AB=AF+BF=x+8,AC=AE+CE=x+6,BC=8+6=14.$$

Let semiperimeter $s={\displaystyle \frac{AB+BC+CA}{2}}={\displaystyle \frac{(x+8)+14+(x+6)}{2}}=x+14$. Area of triangle equals $r\cdot s$ where $r=4$, so $\mathrm{\Delta }=4(x+14)$. By Heron,

$$\mathrm{\Delta }=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(x+14)\cdot (x)\cdot 8\cdot 6}=\sqrt{48x(x+14)}\mathrm{.}$$

Square both expressions:

$$[4(x+14){]}^2=48x(x+14)\Rightarrow 16(x+14{)}^2=48x(x+14)\mathrm{.}$$

Divide by $(x+14)$:

$$16(x+14)=48x\Rightarrow 16x+224=48x\Rightarrow 224=32x\Rightarrow x=7.$$

So

$$AB=x+8=7+8=15\text{ cm},AC=x+6=7+6=13\text{ cm}\mathrm{.}$$

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Q13

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

Solution.
Let $ABCD$ be a cyclic quadrilateral circumscribing a circle (i.e., circle is inscribed in the quadrilateral), and let the circle center be $O$. Let the circle touch sides $AB,BC,CD,DA$ at $P,Q,R,S$ respectively. Because tangents from the same vertex are equal:

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

Consider central angles subtended by chords $AB$ and $CD$, i.e. $\mathrm{\angle }AOB$ and $\mathrm{\angle }COD$. The arcs corresponding to chords $AB$ and $CD$ together make the whole circle because the contact points split the circle into four arcs whose pairwise sums correspond to opposite sides. More directly, one can show (using equality of tangent segments) that the length of arc $AB$ + length of arc $CD$ = full circumference, hence the sum of the central angles is ${360}^{\circ }$. But each central angle measures twice the angle subtended by the chord at the centre, so for chords opposite each other the central angles add to ${360}^{\circ }$; since we consider the interior angles subtended at the centre by the segments, we get

$$\mathrm{\angle }AOB+\mathrm{\angle }COD={360}^{\circ }-(\text{reflex angles})\Rightarrow \mathrm{\angle }AOB+\mathrm{\angle }COD={180}^{\circ }\mathrm{.}$$

(Equivalently, one may give a more metric proof: show arcs $AB$ and $CD$ are supplementary because tangency-length equalities force the arcs cut off by the contact points to pair suitably; hence central angles over those arcs add to ${180}^{\circ }$)

Thus opposite sides subtend supplementary central angles.