Question

The circle $S_{1}$ with centre $C_{1}\left(a_{1}, b_{1}\right)$ and radius $r_{1}$ touches externally the circle $S_{2}$ with centre $C_{2}\left(a_{2}, b_{2}\right)$ and radius $r_{2}$. If the tangent at their common point passes through the origin, then

• A. $ (a_{1}^{2}+a_{2}^{2})+(b_{1}^{2}+b_{2}^{2})=r_{1}^{2}+r_{2}^{2} $
• B. $ (a_{1}^{2}-a_{2}^{2})+(b_{1}^{2}-b_{2}^{2})=r_{1}^{2}-r_{2}^{2} $
• C. $ {{(a_{1}^{2}-{{b}_{2}})}^{2}}+(a_{2}^{2}+b_{2}^{2})=r_{1}^{2}+r_{2}^{2} $
• D. $ (a_{1}^{2}-b_{1}^{2})+(a_{1}^{2}+b_{2}^{2})=r_{1}^{2}+r_{2}^{2} $

Answer 1

Answer: (B)

Given two circles are
$S_{1} \equiv\left(x-a_{1}\right)^{2}+\left(y-b_{1}^{2}\right)=r_{1}^{2} \,\,\,...(i)$
$S_{2} \equiv\left(x-a_{2}\right)^{2}+\left(y-b_{2}^{2}\right)=r_{2}^{2}\,\,\,...(ii)$
The equation of the common tangent of these two circles is given by $S_{1}-S_{2}=0$ i.e.,
$2 x\left(a_{1}-a_{2}\right)+2 y\left(b_{1}-b_{2}\right)+\left(a_{2}^{2}+b_{2}^{2}\right)-\left(a_{1}^{2}+b_{1}^{2}\right)+r_{1}^{2}-r_{2}^{2}=0$
If this passes through the origin, then
$\left(a_{2}^{2}+b_{2}^{2}\right)-\left(a_{1}^{2}+b_{1}^{2}\right)+r_{1}^{2}-r_{2}^{2}=0$
$\Rightarrow \left(a_{2}^{2}-a_{1}^{2}\right)+\left(b_{2}^{2}-b_{1}^{2}\right)=r_{2}^{2}-r_{1}^{2}$
$\Rightarrow \left(a_{1}^{2}-a_{2}^{2}\right)+\left(b_{1}^{2}-b_{2}^{2}\right)=\left(r_{1}^{2}-r_{2}^{2}\right)$