S Chand Class 10 CHAPTER 14 Circle Exercise 14 C

 Exercise 14 C 

Question 1

Ans: (a) In the figure APB is tangent to the circle with center O 

$\angle Q P D=50^{\circ}$

if $O P$ is the radius and $A P B$ is tangent

So $O P \perp A P B$

So $\angle O P B=90^{\circ} \Rightarrow \angle O P Q+\angle Q P B=90^{\circ}$

$\begin{aligned}&\Rightarrow \angle O P Q+30^{\circ}=90^{\circ} \\&\Rightarrow \angle O P Q=90^{\circ}-50^{\circ} \\&\Rightarrow O P Q=40^{\circ}\end{aligned}$

But in $\triangle O P Q, O P=O Q$

So

$\begin{aligned}&\angle O P Q=\angle O Q P=40^{\circ} \\&\angle P O Q+\angle O P Q+\angle O Q P=180^{\circ} \\&\Rightarrow \angle P O Q+40^{\circ}+40^{\circ}=180^{\circ} \\&\Rightarrow \angle P O Q+80^{\circ}=180^{\circ} \\&\text { So } \angle P OQ=180^{\circ}-80=100^{\circ}\end{aligned}$

(b) In circle, two tangent $A B$ and $A C$ are drawn, a point A outside the circle

so $A C=A B=4 \mathrm{~cm}$

(c) In the figure a circle with center O from a point P out side the circle Two tangents PQ and PR are 

Drawn and $\angle Q P R=80^{\circ}$

So $\angle Q P R$ and $\angle Q O R$ are supplementry

So $\angle Q P R+\angle Q O R=180^{\circ}$

$\begin{aligned}&\Rightarrow 80^{\circ}+\angle Q O R=180^{\circ} \\&\Rightarrow \angle Q O R=180^{\circ}-80^{\circ}=100^{\circ} \\&\text { SO } \angle Q O R=100^{\circ}\end{aligned}$

Question 2

Ans: In the figure a circle with center 0 from a point $P$ outside of it , tangents $P T$ and $P S$ are drawn to the circle and $\angle T P O=30^{\circ}$

In $\triangle P T O, O T \perp P T$

So $\angle O T P=90^{\circ}$

So $\angle T O P+\angle T P O=90^{\circ}$

$\begin{aligned}&\Rightarrow \angle \text { TOP }+30^{\circ}=90^{\circ} \\&\Rightarrow \angle T O P=90^{\circ}-30^{\circ}=60^{\circ}\end{aligned}$

So OP is the bisects of ㄥTOS

So $\angle T O P=\angle P O S=60^{\circ}$

Question 3

Ans: In the figure $B D$ is the diameter of the circle $P Q$ it tangent to the circle at $A$

$\angle A D B=30^{\circ}, \angle O B C=60^{\circ}$

(i) If QAP is tangent and AB is chord of the circle 

So  $\angle Q A B=\angle A D B=30^{\circ}$

(ii) $\angle P A D+\angle D A B+\angle Q A B=180^{\circ}$

$\Rightarrow \angle P A D+90^{\circ}+30^{\circ}=180^{\circ}$

$\begin{aligned}&\Rightarrow \angle P A D+120^{\circ}=180^{\circ} \\&\Rightarrow \angle P A D=180^{\circ}-120^{\circ}=60^{\circ}\end{aligned}$

(iii) In $\triangle B C D$

$\begin{aligned}& \angle C O B+\angle C B D+\angle B C D=180^{\circ} \\\Rightarrow & \angle C D B+60^{\circ}+90^{\circ}=180^{\circ} \\\Rightarrow & \angle C D B+150^{\circ}=180^{\circ} \\\Rightarrow & \angle C D B=180^{\circ}-150^{\circ}=30^{\circ}\end{aligned}$

Question 4

Ans: In the figure PQ and PR the tangents drawn from P outside the circle such that 

PQ= PR =9cm and $\angle Q P R=60^{\circ}$

$Q R$ is joined

In $\triangle P Q R 1<Q P R=60^{\circ}$

if $P Q=P R$

So $\angle P Q R=\angle P R Q=60^{\circ}$

So $\triangle P Q R$ is an equilateral triangle $P Q=P R=Q R=9 \mathrm{~cm}$

Question 5

Ans: In a circle of radius 3cm, point P is 3cm  away from the center O of the circle 

PQ and PR are the tangents drawn from P to the circle 

(IMAGE TO BE ADDED)

if OQ is radius and PQ is tangent

So $O Q \perp Q P$ or $\angle O Q P=90^{\circ}$

Now in right angled $\triangle O P Q$.

$\begin{aligned}&O P^{2}=O Q^{2}+P Q^{2} \\&=(5)^{2}=(3)^{2}+P Q^{2}\end{aligned}$

$\Rightarrow 25=9+P Q^{2} \Rightarrow P Q^{2}=25-9=16=(4)^{2}$

So $P Q=4 \mathrm{~cm}$

BUT PQ $=P R$

So $P Q=P R=4 \mathrm{~cm}$

Question 6

Ans:  (IMAGE TO BE ADDED)

if from A, A Q and A R the tangents drawn to the circle

So $\quad A Q=A R=5 \mathrm{~cm}$.......(i)

Similarly from B tangent BQ and BP are drawn 

So $B Q=B P$...........(ii)

and from C

C P=C R

 Now perimeter of $\triangle A B C$,

$=A B+A C+B C=A B+A C+B P+C P$

$=A B+A C+B Q+C R$

$=A B+B Q+A C+C R$

$=A Q+A R=5 \mathrm{~cm}+5 \mathrm{~cm}$

$=10 \mathrm{~cm}$

Question 7

Ans:  Two circle, which are concentric and their center is O, are radii 5cm and 3cm 

i.e OA = 5cm and OP = 3cm 

AB is chord of larger circle which touches the smaller circle at P 

Question 8

Ans: $A B C$ is a triangle and with center $A \perp B$ and $C$, three circle are drawn touching each other externally at $P, Q$ and $R$ respectively $A B=4 \mathrm{~cm}, B C=7 \mathrm{~cm}$ and $A C=6 \mathrm{~cm}$

(IMAGE TO BE ADDED)

Let radii of circle with center $A, B$ and $C$ respectively be $x, y$ and $z$

So $A B=x+y, B C=y+2, C A=2+x$

$\begin{gathered}\Rightarrow x+y=4 \mathrm{~cm}, y+z=6 \mathrm{~cm}, z+x=7 \mathrm{~cm} \\\text { So } A B+B C+C A=x+y+y+z+z+x \\\Rightarrow 4+6+7=2(x+y+z) \\\Rightarrow x+y+z=\frac{17}{2}=8.5 \mathrm{~cm}\end{gathered}$

Subtracting From $x+y+z$, we qut

$x=8.5-4=4.5 \mathrm{~cm}$

$y=8.5-6=2.5 \mathrm{~cm}$

$z=8.5-7=1.5 \mathrm{~cm}$

Hence their radii are $2.5 \mathrm{~cm} 1.5 \mathrm{~cm}$ and $4.5 \mathrm{~cm}$

Question 9

Ans: Two equal circles with center O and O' touch each other externally at X. OO' is produce to meet the circle O' at A. Through A, a tangent AC is drawn to the circle with center O. O'D$\perp A C$

Let $r$ be the radius of each circle 

In $\triangle A O^{\prime} D$ and $\triangle A O C_{1}$

$\angle D=\angle C$

$\angle A=\angle A$

(i) So $\triangle A O^{\prime} O \sim \triangle A O C$

So $\frac{A O^{\prime}}{A O}=\frac{r}{A X+X 0}=\frac{r}{2 r+r}=\frac{r}{3 r}=\frac{1}{3}$

(ii) So

$\begin{aligned}\frac{\text { area of } \triangle A D O^{\prime}}{\text { area of } \triangle A C O}=\frac{A O^{\prime2}}{A O^{2}} &=\left(\frac{1}{3}\right)^{2} \\&=\frac{1}{9}\end{aligned}$

Question 10

Ans: Two circles with centers $P$ and $Q$ touch externally at $R$ $\times 4$ is their common tangent

$A$ and $B$ are their points af contact

Join $P A, Q B$ and $P Q$

from $Q$, draw QS|| XY

(IMAGE TO BE ADDED)

if $P A$ and $Q B$ are perpendicular to $x 4$ and $Q S$ || $A B$

So $Q S=A B$

$P A=12 \mathrm{~cm}, Q B=3 \mathrm{~cm}, P Q=12+3=15 \mathrm{~cm}$

So $P S=P A-S A=12-3=9 \mathrm{~cm}$

Now in right $\triangle P S Q$,

$P Q^{2}=P S^{2}+Q S^{2}$

$=(15)^{2}=(9)^{2}+Q S^{2} \Rightarrow 225=81+Q S^{2}$

$\Rightarrow Q S^{2}=225-81=144=(12)^{2}$

So $Q S=12 \mathrm{~cm}=A B=Q S=12 \mathrm{~cm}$