The equation of the circle which is orthogonal to the circles $x^2+y^2+2x+17y+4=0$, $x^2+y^2+7x+6y+11=0$, $x^2+y^2-x+22y+3=0$ is

The circle with centre as radical centre of three circles and having radius equal to the length of tangent from it to any circle cuts the three circles orthogonally.By solving radical axes Radical centre is (3, 2) and length of tangent is $\sqrt{57}$. The equation of required circle is 

$(x-3{)}^2+(y-2{)}^2=(\sqrt{57}{)}^2$

$x^2+y^2-6x-4v+9+4-57=0$

$\text{ i.e., }{x}^2+y^2-6x-4y-44=0$