Question

Find the equation of the system of coaxial circles that are tangent at $(\sqrt{2},4)$ to the locus of the point of intersection of two mutually perpendicular tangents to the circle $x^2+y^2=9$.

Answer 1

Locus of the point of intersection of two mutually perpendicular tangents to the circle is the director circle,

eq. of director circle=$x^2+y^2=(r\sqrt{2}{)}^2$

eq. of director circle of given circle = $x^2+y^2=18$

$Replacexby(x-a)andyby(y-b)$

equation of the system of coaxial circles is $(x-\sqrt{2}{)}^2+(y-4{)}^2=18$