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.,
$2x\left(a_1-a_2\right)+2y\left(b_1-b_2\right)+\left(a_2^2+b_2^2\right)-\left(a_1^2+b_1^2\right)+r_1^2-r_2^2=0$
If this passes through the origin, then
$\left(a_2^2+b_2^2\right)-\left(a_1^2+b_1^2\right)+r_1^2-r_2^2=0$
$\Rightarrow \left(a_2^2-a_1^2\right)+\left(b_2^2-b_1^2\right)=r_2^2-r_1^2$
$\Rightarrow \left(a_1^2-a_2^2\right)+\left(b_1^2-b_2^2\right)=\left(r_1^2-r_2^2\right)$