Question

If two circles $x^2+y^2-2ax+c^2=0$ and $x^2+y^2-2by+c^2=0$ touch each other externally, then

• A. $\frac{1}{a^2}+\frac{1}{c^2}=\frac{1}{b^2}$
• B. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
• C. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{2}{c^2}$
• D. $\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}$

Answer 1

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

Let $S_1:x^2+y^2-2ax+c^2=0$
$C_1\equiv (a,0)$ and $r_1=\sqrt{a^2-c^2}$
and $S_2:x^2+y^2-2by+c^2=0$
$C_2\equiv (0,b)$ and $r_2=\sqrt{b^2-c^2}$

$C_1{C}_2=\sqrt{(a-0{)}^2+(0-b{)}^2}$
$\Rightarrow C_1{C}_2=\sqrt{a^2+b^2}$
and $r_1+r_2=\sqrt{a^2-c^2}+\sqrt{b^2-c^2}$
Since the two circles touch each other externally, we have
$C_1{C}_2=r_1+r_2$
$\Rightarrow \sqrt{a^2+b^2}=\sqrt{a^2-c^2}+\sqrt{b^2-c^2}$
$\Rightarrow a^2+b^2=a^2-c^2+b^2-c^2+2\sqrt{a^2-c^2}\sqrt{b^2-c^2}$
$\Rightarrow c^2=\sqrt{a^2-c^2}\sqrt{b^2-c^2}$
$\Rightarrow c^4=a^2{b}^2-c^2(a^2+b^2)+c^4$
$\Rightarrow a^2{b}^2=c^2(a^2+b^2)$
$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$