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
Best Answer
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