Question

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

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

Answer 1

The correct option is A

$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}$

Given : $x^2+y^2+2ax+c=0\dots \left(1\right)$
And $x^2+y^2+2by+c=0\dots \left(2\right)$

For circle (1), we have :
Centre $=(-a,0)=C_1$

For circle (2), we have :
Centre $=(0,-b)=C_2$

Let the circles intersect at point P.

$\therefore$ Coordinates of P = Mid point of $C_1{C}_2$

$\Rightarrow$ Coordinates of

$P=(\frac{-a+0}{2},\frac{0-b}{2})=(\frac{-a}{2},\frac{-b}{2})$

Now, we have:

$P{C}_1$ = radius of (1)

$\Rightarrow \sqrt{{(-a+\frac{a}{2})}^2+{(-0-\frac{b}{2})}^2}=\sqrt{a^2-c}$

$\Rightarrow \frac{a^2}{4}+\frac{b^2}{4}=a^2-c\dots \left(3\right)$

Also, radius of circle (1)=radius of circle (2)

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

$\Rightarrow a^2=b^2\dots \left(4\right)$

From (3) and (4), we have:

$\frac{a^2}{2}=a^2-c$

$\Rightarrow \frac{2}{a^2}=\frac{1}{c}$

$\Rightarrow \frac{1}{a^2}+\frac{1}{a^2}=\frac{1}{c}$

$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}$