Question

Find the equation of the radical axis of the circles $2x^2+2y^2+3x+6y-5=0$ and $3x^2+3y^2-7x+8y-11=0$.

Answer 1

Given equation of circles-

$S_1:2x^2+2y^2+3x+6y-5=0$

$\Rightarrow S_1:x^2+y^2+\frac{3}{2}x+3y-\frac{5}{2}=0$

$S_2:3x^2+3y^2-7x+8y-11=0$

$\Rightarrow S_2:x^2+y^2-\frac{7}{3}x+\frac{8}{3}y-\frac{11}{3}=0$

As we know that equation of radical axis of two circles is given as-

$S_1-S_2=0$

$(x^2+y^2+\frac{3}{2}x+3y-\frac{5}{2})-(x^2+y^2-\frac{7}{3}x+\frac{8}{3}y-\frac{11}{3})=0$

$\Rightarrow (\frac{3}{2}+\frac{7}{3})x+(3-\frac{8}{3})y-(\frac{5}{2}-\frac{11}{3})=0$

$\Rightarrow \frac{23}{6}x+\frac{1}{3}y+\frac{7}{6}=0$

$\Rightarrow 23x+2y+7=0$

Hence the equation of radical axis of the given circles is $23x+2y+7=0$