Question

Find the radical axis of the pairs of circles
$x^2+y^2-3x-4y+5=0$ and $3x^2+3y^2-7x+8y+11=0$.

Answer 1

The equation of the radical axis for a pair of circles $S_1$ and $S_2$ is given by
$S_1-S_2=0$.

In the above case $S_1=0\Rightarrow x^2+y^2-3x-4y+5=0$

and $S_2=0\Rightarrow 3x^2+3y^2-7x+8y+11=0$.
Now $S_2$ can be re-writen as

$x^2+y^2-(\frac{7}{3}x+\frac{8y}{3}-\frac{11}{3})=0$

The radical axis of two circles be

$S_1-S_2=0$

Hence,

$\frac{-2}{3}x-\frac{20}{3}y+\frac{4}{3}=-2x-20y+4=0=x+10y-2=0$.

Hence the equation of the radical axis is $x+10y-2=0$.