Question

There are two circles $C_{1}$ and $C_{2}$ whose radii are $r_{1},r_{2}$ , respectively. If distance between their centre is $3r_{1}-r_{2}$ and length of direct common tangent is twice of the length of transverse common tangent. Then $r_{1}:r_{2}$ is

• A. $5:4$
• B. $6:5$
• C. $7:6$
• D. $8:7$

Answer 1

Answer: (C)

Length of direct common tangent $=\sqrt{d^{2} - \left(r_{1} - r_{2}\right)^{2}}$
Length of transverse common tangent $=\sqrt{d^{2} - \left(r_{1} + r_{2}\right)^{2}}$
Where $d$ is distance between their centre. Then
$\left(\sqrt{d^{2} - \left(r_{1} - r_{2}\right)^{2}}\right)=2\left(\sqrt{d^{2} - \left(r_{1} + r_{2}\right)^{2}}\right)$
$d^{2}=\left(r_{1}^{2} + r_{2}^{2} + \frac{10}{3} r_{1} r_{2}\right),$ where $d=3r_{1}-r_{2}$
$\left(3 r_{1} - r_{2}\right)^{2}=r_{1}^{2}+r_{2}^{2}+\frac{10}{3}r_{1}r_{2}$
$r_{1}:r_{2}=7:6$