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 $BA_3$ 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 $BA_1 = 2, BA_2 = 3$ and $BA_3 = 4$, 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 $C_1$.

$BA_1 = 2$
$BA_2 = 3$
$BA_3 = 4$
Let radius of circle $C_1 = r_1$ and radius of circle $C_2 = r_2$
$\therefore BA = 2r_1$ and $AC = 2r_2$
$\Rightarrow BM = \frac{1}{2} BA_1 = 1$
$\Rightarrow BN = BA_2 + \frac{1}{2} A_2A_3$
$ = 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, $BA_2 \times BA_3 = BA \times BC$
$\Rightarrow 3 \times 4 = 2r_1 (2r_1 + 2r_2)$
$\Rightarrow 12 = 4 (r^2_1 + r_1r_2)$
$\Rightarrow r^2_1 + r_1r_2 = 3\,...(ii)$
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}$