Question

The circles whose equations are $x^2+y^2+c^2=2ax$ and $x^2+y^2+c^2-2by=0$ will touch each other externally if

• A. $\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}$
• B. $\frac{1}{c^2}+\frac{1}{a^2}=\frac{1}{b^2}$
• C. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$

The two circles are
$x^2+y^2-2ax+c^2=0$
and $x^2+y^2-2by+c^2=0$
centres: $C_1(a,0)$ $C_2(0,b)$
radii: $r_1=\sqrt{a^2-c^2}$, $r_2=\sqrt{b^2-c^2}$
Since, the two circles touch each other externally
Therefore, $C_1{C}_2=r_1+r_2$
$\Rightarrow \sqrt{a^2+b^2}=\sqrt{a^2-c^2}+\sqrt{b^2-c^2}$
$\Rightarrow a^2+b^2=a^2-c^2+b^2-c^2+2\sqrt{a^2-c^2}\sqrt{b^2-c^2}$
$\Rightarrow c^4=a^2{b}^2-c^2(a^2+b^2)+c^4$
$\Rightarrow a^2{b}^2=c^2(a^2+b^2)$
$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$