Question

The perpendicular distance from the origin to the radical axis of the circles $2x^2+2y^2-3x-y+3=0$ and $3x^2+3y^2-x+y-1=0$ is

• A. $\sqrt{2}$
• B. $\frac{11}{\sqrt{74}}$
• C. $\frac{\sqrt{5}}{2}$
• D. $\sqrt{\frac{5}{2}}$

Answer 1

Answer: (B)

$\frac{11}{\sqrt{74}}$

$s_1:2x^2+2y^2-3x-y+3=0$
$s_2:3x^2+3y^2-x+y-1=0$
So, equation of radial axis is
$s_1-s_2=0$
$(-9x-3y+9)-(-2x+2y-2)=0$
$-7x-5y+11=0$
$\Rightarrow 7x+5y-11=0$
Perpendicular distance from $P(0,0)$ to $7x+5y-11=0$ is
$d=|\frac{11}{\sqrt{74}}|=\frac{11}{\sqrt{74}}$