Question

The centre of circle $S$ lies on the line $2x-2y+9=0$ and $S$ cuts at right angles the circle $x^2+y^2=4$, show that $S$ passes through each of two fixed points and find their co-ordinates.

Answer 1

Let the circle be
$x^2+y^2+2gx+2fy+c=0$
Its centre $(-g,-f)$ lies on $2x-2y+9=0$.
$\therefore -2g+2f+9=0.....\left(1\right)$
It cuts orthogonally the circle $x^2+y^2-4=0$.
$\therefore 2g_1{g}_2+2f_1{f}_2=c_1+c_2$
or $0+0=c-4\therefore c=4$
and from $\left(1\right),2g=9+2f$.
Putting the values of $c$ and $g$ we get the equation of the circle as
$x^2+y^2+(9+2g)x+2fy+4=0$
or $(x^2+y^2+9x+4)+2f(x+y)=0$.
Above is of the form $S+\lambda P=0$
Above represents a family of circles which passes through the points of intersection of $S=0$ and $P=0$.
Solving $x^2+y^2+9x+4=0$ and $x+y=0$ we get the fixed points as $(-1/2,1/2)$ and $(-4,4)$.