Finding the intersection points of 2 circles with a given equation, center and radius Determine the points of intersection of the circle with equation $x^2 + y^2 - 12x - 4y + 30 = 0$ and the circle with center $(3,5)$ and radius $4$.
My attempt :
With given center and radius of the second circle, I make an equation:
$(x - a)^2 + (y-b)^2 = r^2$ 
$(x - 3)^2 + (y-5)^2 = 4^2$
$ x^2 -6x +9 + y^2 -10y +25 = 16 $
$ x^2 + y^2 - 6x -10y +18 = 0 $
And then I equate both equations:
$ x^2 + y^2 - 6x -10y +18 = x^2 + y^2 - 12x - 4y + 30 $
and at the end it gives me:
$-x + y = -2$
Am I doing it in the right way? what should I do next?

So you have the points of intersection of the circles satisfy $y = x-2$.  Using this in any one circle gives you the points. e.g.
$$(x-3)^2+(x-7)^2=16 \implies x = 3,7$$
