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

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

## 2 Answers

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

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.