Question

The two circles $x^2+y^2+2ax+c=0and{x}^2+y^2+2by+c=0$ touch if $\frac{1}{a^2}+\frac{1}{b^2}=$

• A. $\frac{1}{c^2}$
• B. $c$
• C. $\frac{1}{c}$
• D. $c^2$

Answer 1

The correct option is C

$\frac{1}{c}$

We can write the equation of common tangent, with the help of radical axis.
Since two circles are touching each other the common tangent will be the radical axis, and the equation of the radical axis we can write -
S = S'
$x^2+y^2+2ax+c=x^2+y^2+2by+c=0$
So, Equation of Common tangent is ax – by = 0.

Now, we can find the perpendicular distance from the center of the first circle to the tangent which also should be equal to radius of that circle.
Perpendicular distance from (-a,0) to the tangent = $\frac{a^2}{\sqrt{a^2+b^2}}$
Radius of the same circle will be = $\sqrt{a^2-c}$
Therefore $\frac{a^2}{\sqrt{a^2+b^2}}=\sqrt{a^2-c}$