Question

Find the equation of the circle which cuts the following circles orthogonally.
$x^2+y^2+2x+4y+1=0,$
$2(x^2+y^2)+6x+8y-3=0,$
$x^2+y^2-2x+6y-3=0$

Answer 1

The Circle having Center at the radical center of the Three given circles and radius equal to the length of the tangent from it to any one of the three circles cuts all three circles orthogonally

Given Circles are $S=x^2+y^2+2x+4y+1=\mathrm{0..................}\left(1\right)S_1=2(x^2+y^2)+6x+8y-3=\mathrm{0..................}\left(2\right)S_2=x^2+y^2-2x+6y-3=\mathrm{0..................}\left(3\right)$

For circles $S-S_1=0,S-S_2=0$

The radical axes of (1),(2),(3) are respectively

$x-\frac{5}{2}=010x-4y+3=0$

From solving above equations we get,

$x=\frac{5}{2},y=7$

Thus, the Coordinates of Radical Circle are $\frac{5}{2},7$

Length of Tangent from $(\frac{5}{2},7)$ to (1)

$r=\sqrt{\frac{25}{4}+49+5+28+1}=\frac{\sqrt{357}}{2}$

Hence, Required Circle is $(x-\frac{5}{2}{)}^2+(y-7{)}^2=\frac{357}{4}$