S Chand Class 10 CHAPTER 14 Circle Exercise 14 D

 Exercise 14 D

Question 1

Ans: In a circle with center O, from a paint P outside of it, PA and PB are the tangents to the circle 

At a point m on the circle another tangent is drawn which intersects AP at K and at N. 

To prove: KN = AK+BN

Proof: From K, KA and KM are tangents to the circle 

So KA=KM ........(i)

Similarly from $N, N M$ and $N B$ tangents are drawn

So $N B=m N$

Adding i and ii

$\begin{aligned}KM+M N &=k A+N B \\k N &=A K+B N\end{aligned}$

Question 2

Ans:  Two concentric circles with center O .

$A$ chord $A B$ is drawn to the greater circle which touches the smaller circle at $P$.

To prove: $A B$ is bisected at $P$ i.e. $A P=P B$ construction: Join DP, OA and OB.

(IMAGE TO BE ADDED)

Proof: if $O P$ is the radiues and $A B$ is the tangent

So OP $\perp A B$

In right $\triangle O A P$ and $\triangle O B P$

HyP. $O A=O B$

Side $O P=O P$

So $\triangle O A P \equiv \triangle O B P$

So $A P=P B$

Hence $P$ bisects $A B$ Hence Proved

Question 3

Ans: A circle with center O, from a point P outside of it, two tangents PT and PS are drawn TS is 

Joined which subtends 

ㄥTOS at the centre

To prove: $\angle T P S+\angle T O S=180^{\circ}$

Construction: Join OP.

(IMAGE TO BE ADDED)

Proof: In right angle $\Delta O T P$

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

So $\angle T O P+\angle T P O=90^{\circ}$.........(i)

Similarly in right $\triangle O S P$

$\angle S O P+\angle S P O=90^{\circ}$.......(ii)

Adding (i) and (ii)

$\begin{aligned}&\angle T O P+\angle S O P+\angle T P O+\angle S P O=90^{\circ}+90^{\circ} \\&\Rightarrow \angle T P S+\angle T O S=180^{\circ}\end{aligned}$

So LTPS and ㄥTOS are supplementary hence proved 

Question 4

Ans: In the figure a circle touches the side BC of a $\triangle A B C$ externally and AB and AC on producing at Y and Z

To prove: $A Y=\frac{1}{2}(A B+B C+C A)$

Proof: if from $A, A Y$ and $A Z$ are tangents to the circle

So $A Y=A Z$

similarly from $B, B y$ and $B x$ are tangents So $B y=B x$

and from CX and CZ are tangents

 So C Z=C X

Now $A Y=A B+B Y=A B+B \times$

$\begin{aligned}&A Z=A C+C Z=A C+C X \\&\text { So } A Y+A Z=A B+B \times+C \times+A C \\&\Rightarrow A Y+A Y=A B+B C+C A \\&\Rightarrow 2 A Y=A B+B C+C A \\&\Rightarrow A Y=\frac{1}{2}(A B+B C+C A)\end{aligned}$

Hence proved 

Question 5

Ans: In circle touches the sides BC , CA and AB at D,E and F respectively of the $\triangle A B C$

To prove: 

$\begin{aligned} A F &+B D+C E=A E+C D+B F \\ &=\frac{1}{2}(A B+B C+C A) \end{aligned}$

Proof: from $A, A F$ and $A E$ are the tangents to the circle

So $A F=A E$.......(i)

Similarly From B, BD and BF are the tangents

So $B D=B F$......(ii)

and From C, CE and $C D$ are the tangents

So $C E=C D$.........(iii)

Adding (i) (ii) and (iii) we get,

$A F+B D+C E=A E+B F+C D$

Adding to both sides

A F+B D+C E

$2 A F+2 B D+2 C E=A E+B F+C D+A F+B D+C E$

$2(A F+B D+C E)=A E+C E+B D+C D+A F+B F$

$\Rightarrow 2(A F+B D+C E)=A C+B C+A B$

$\Rightarrow A F+B D+C D=\frac{1}{2}(A B+B C+C A)$ Hence proved 

Question 6

Ans: Two circles touch each other externally at $T$.

$P T$ is their common tangent and from $P$, tangent $P A$ and $P B$ are drawn to the two circles

To prove: $P A=P B$

proof: if from P,PA and PT are the tangents to the bigger circle

So $\quad P A=P T$...........(i)

Similarly from $P, P T$ and $P B$ are the tangents drawn to the smaller circle

So $P T=P B$........(ii)

from (i) and (ii)

P A=P B Hence proved 

Question 8

Ans: Two circles touches each other at A externally. Through A, a common tangent is drawn. A point $P$ is taken on the tangents.

(IMAGE TO BE ADDED)

To prove : $P B=P C$

-Proof: Through $P_{1}$ two tangents $P A$ and $P B$ and drawn to the first circle So P A=P B......... (i)

Again through P, AB and $P C$ are the tangents drawn to the second circle

So $\quad P A=P C$

From (i) and (ii)

PB= PC Hence proved 

Question 9

Ans: In the circle with center 0 , from a point $p$ outside of it  PA and PB are tangents drawn $O P, O A, O B$ and $A B$ are joined

To prove:

(i) $\angle A O P=\angle B O P$

(ii) OP is the perpendicular bisector of AB

 proof:

(i) In right $\triangle O A P$ and $\triangle O B P$.

Hyp OP =OP

side $O A=O B$

So $\triangle O A P \cong \triangle O B P$

So $\angle A O P=\angle B O P$

and $\angle A P O=\angle B P O$

Now in $\triangle A M P$ and $\triangle B M P$

$MP=MP$

$P A=P B$

$\angle A P M=\angle B P M$

So $\triangle A M P \cong \triangle B M P$

So $A M=B M$

$\angle A M P=B M P$

But $\angle A M P+\angle B M P=180^{\circ}$

So $\angle A M P=\angle B M P=90^{\circ}$

Hence OP is the perpendicular bisects of AB 

Question 10

Ans: In a circle with center O, AB is the chord 

Through A and B tangents PQ and PR are drawn which meet each other at P

(IMAGE TO BE ADDED)

To prove: $\angle P A B=\angle P B A$

construction: Join $O A, O B$ and $O P$

Proof: In $\triangle O A P$ and $\triangle O B P$,

$O P=O P$

$O A=O B$

$P A=P B$

So $\triangle O A P \cong \triangle O B P$

So $\angle A P O=\angle B P O$

Now in $\triangle P A M$ and $\triangle P B M$,

$\begin{aligned}&P M=P M \\&P A=P B\end{aligned}$

$\angle A P M=\angle B P M ( A P O=\angle B P O)$

So $\triangle P A M \cong \triangle P B M$

So $\angle P A M=\angle P B M$

$\Rightarrow \angle P A B=\angle P B A$

Question 11

Ans: Two circles with centers O and O' touch each other externally at C

$P Q$ and $R S$ are common tangents and a third common tangent from $C$ is drawn which intersecting $P Q$ and $R S$ and $L$ and M

To prove: LM bisects PQ and RS 

Proof: From L, LP and LC are the tangents to the circle 

So LP = LC .........(i)

Similarly from  and L, LQ and LC are two tangents 

SO LC =L Q$

from (i) and (ii)

LP = LC =LQ 

So L is the mid point of PQ 

Similarly we can prove that 

$m$ is the mid point of RS

Hence $L m$ bisects the tangents $P Q$ and RS

Question 12

Ans:  A circle with center C . $P A$ and $P B$ are two tangents drawn from $P$ to the circle $C A$ is joined.

To prove: $\angle A P B=2 \angle C A B$

Proof: In $\triangle A P B$,

$A P=B P$

So $\angle P A B=\angle P B A$

But $\angle P A B+\angle P B A+\angle A P B=180^{\circ}$

$\begin{aligned}\Rightarrow & \angle P A B+\angle P A B+\angle A P B=180^{\circ} \\\Rightarrow & \angle A P B=180^{\circ}-2 \angle P A B \\& \text { But } \angle P A B=\angle C A P-\angle C A B \\&=90^{\circ}-\angle C A B \\\text { So } \angle A P B=180^{\circ}-2\left(90^{\circ}-\angle C A B\right) \\\Rightarrow & \angle A P B=180^{\circ}-180^{\circ}+2 \angle C A B \\\Rightarrow & \angle A P B=2 \angle C A B\end{aligned}$

Hence proved