Question

Circle $C_1$ has radius $2$ and circle $C_2$ has radius $3$, and the distance between the centers of $C_1$ and $C_2$ is $7$. If two lines, one tangent to both circles and other passing through the center of both circles, intersect at point $P$ which lies between centers of $C_1$ and $C_2$, the the distance between $P$ and the center of $C_1$

• A. $\frac{9}{4}units$
• B. $\frac{7}{3}units$
• C. `$\frac{8}{3}units$
• D. $\frac{14}{5}units$

Answer 1

The correct option is D $\frac{14}{5}units$

REF.Image.

Given : $C_{1}M=2cm$ and $N{C}_2=3cm$

$\angle C_{1}AB=\angle C_{2}BA={90}^{\circ }$

And $C_1{C}_2=7cm$

To Find :$C_{1}P$

$C_1{C}_2=C_{1}M+MN+N{C}_2$

$MN=2cm$

Let MP be x cm, so NP = 2-x

In $△C_{1}AP$ and $△C_{2}BP$

$\angle C_{1}AP=\angle C_{2}BP={90}^{\circ }$ [Given]

$C_{1}PA=\angle C_{2}PB$ [vertically opposite angles]

$△C_{1}AP\sim △C_{2}BP$ [by AA similarly]

$\frac{C_{1}A}{C_{2}B}=\frac{AP}{BP}=\frac{C_{1}P}{C_{2}P}$

$\frac{2}{3}=\frac{2+x}{5-x}$

$10-2x=6+3x$

$5x=4\Rightarrow x=\frac{4}{5}$

$C_{1}P=C_{1}M+NP$

$=2+\frac{4}{5}=\frac{14}{5}cm$ Ans