Question

Let $C_{1}$ and $C_{2}$ be two circles of radius $r_{1}$ and $r_{2}$ respectively $\left(r_{1} > r_{2}\right)$ touching both the axes. If the two circles are orthogonal, then $\frac{r_{1}}{r_{2}}$ is equal to

• A. $2$
• B. $2+\sqrt{3}$
• C. $3+\sqrt{2}$
• D. $3$

Answer 1

Answer: (B)

Let the equation of a circle touching both the axes be
$\left(x - r\right)^{2}+\left(y - r\right)^{2}=r^{2}$
$x^{2}+y^{2}-2rx-2ry+r^{2}=0$
If the two given circles are orthogonal, then
$2\left(- r_{1}\right)\left(- r_{2}\right)+2\left(- r_{1}\right)\left(- r_{2}\right)=r_{1}^{2}+r_{2}^{2}$
$\Rightarrow r_{1}^{2}-4r_{1}r_{2}+r_{2}^{2}=0$
$\Rightarrow \frac{r_{1}}{r_{2}}=2+\sqrt{3}$