Question

If the circles $x^2+y^2+2ax+c=0$ and $x^2+y^2+2by+c=0$ touch each other then

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

Answer 1

The correct option is A $a^{-2}+b^{-2}=c^{-1}$

Given circles are $C_1:x^2+y^2+2ax+c=0$ and $C_2:x^2+y^2+2by+c=0$.

$C_1(-a,0)$ and $r_1=\sqrt{a^2-c}$

$C_2(-b,0)$ and $r_1=\sqrt{b^2-c}$

Since, circles touch each other

$C_1{C}_2=r_1+r_2$

$|a-b|=\sqrt{a^2-c}+\sqrt{b^2-c}$

$a^2+b^2-2ab=a^2-c+b^2-c+2\sqrt{a^2-c}\sqrt{b^2-c}$

$c-ab=\sqrt{a^2{b}^2-c(a^2+b^2)+c^2}$

$a^{-2}+b^{-2}=c^{-1}$
Ans: A