Question

Let $C_1,C_2$ be two circles touching each other externally at the point $A$ and let $AB$ be the diameter of circle $C_1$ Draw a secant $B{A}_3$ to circle $C_2$, intersecting circle $C_1$ at a point $A_1(\ne A)$, and circle $C_2$ at points $A_2$ and As. If $B{A}_1=2,B{A}_2=3$ and $B{A}_3=4$, then the radii of circles and $C_2$ are respectively

• A. $\frac{\sqrt{30}}{5},\frac{3\sqrt{30}}{10}$
• B. $\frac{\sqrt{5}}{2},\frac{7\sqrt{5}}{10}$
• C. $\frac{\sqrt{6}}{2},\frac{\sqrt{6}}{2}$
• D. $\frac{\sqrt{10}}{3},\frac{17\sqrt{10}}{30}$

Answer 1

Answer: (A)

Given,
$AB$ is diameter of circle $C_1$.

$B{A}_1=2$
$B{A}_2=3$
$B{A}_3=4$
Let radius of circle $C_1=r_1$ and radius of circle $C_2=r_2$
$\therefore BA=2r_1$ and $AC=2r_2$
$\Rightarrow BM=\frac{1}{2}B{A}_1=1$
$\Rightarrow BN=B{A}_2+\frac{1}{2}{A}_2{A}_3$
$=3+\frac{1}{2}=\frac{7}{2}$
In $\Delta BMP$ and $\Delta BNQ$,
$\Delta BMP\sim \Delta BNQ$
$\therefore \frac{BM}{BN}=\frac{BP}{BQ}$
$\Rightarrow \frac{1}{7/2}=\frac{r}{2r_1+r_2}$
$\Rightarrow 2r_2=2r_1...(i)$
Now, $B{A}_2\times B{A}_3=BA\times BC$
$\Rightarrow 3\times 4=2r_1(2r_1+2r_2)$
$\Rightarrow 12=4(r_1^2+r_1{r}_2)$
$\Rightarrow r_1^2+r_1{r}_2=3...(ii)$
From Eqs. (i) and (ii), we get
$r_1=\sqrt{\frac{6}{5}}=\frac{\sqrt{30}}{5}$ and $r_2=\frac{3\sqrt{30}}{10}$