Finding fixed points the circle $x^2+y^2+2gx+(2g-9)y+4=0$ passes through

Question

I am supposed to find two pairs of points the circle meeting the following constraints passes through.

A circle cutting $x^2+y^2=4$ orthogonally and having its center on the line $2x-2y+9=0$ passes through two fixed points. Find them.

---

My Attempt

Let the circle be $x^2+y^2+2gx+2fy+c=0$ where $g,f,c\in \mathbb{R}$. Since the circle cuts $x^2+y^2=4$ orthogonally. So by using the condition for orthogonality we get that $c=4$. Now using the fact that the centre of the circle lies on $2x-2y+9=0$, we have $-2g+2f+9=0$. So the equation of the circle can be written as $x^2+y^2+2gx+(2g-9)y+4=0$.

---

I do not know how to proceed. Any hints would be appreciated. Kindly also inform about the techniques for finding fixed points for any curve. Thanks

Answer 1

You have done well so far. Now, write your final expression as $$x^2+y^2-9y+4 + g(2x+2y)=0$$ If you observe, this expression represents a family of circles which pass through the point of intersection of the circle $x^2+y^2-9y+4 =0$ and the line $x+y=0$. Therefore the fixed points are obtained when you solve the equations of the line and circle simultaneously which is quite simply obtained by plugging $x=-y$ in the equation of the circle

Answer 2

Your equation can be rewritten as: $$ x^2+y^2+2g(x+y)-9y+4=0. $$ As we seek solutions valid for any value of $g$, that is possible only if $x+y=0$. Plug then $y=-x$ in the equation to find the solutions.