# Solve this system of equations by hand: $(x-6)^2 + y^2 = 50, x^2 + (y+2)^2 = 50$

algebra-precalculus

I'm having trouble solving the following system of equations by hand:

$$(x-6)^2 + y^2 = 50 \\x^2 + (y+2)^2 = 50$$

I've tried expanding and removing the square terms, but then I'm left with 2 unknown linear terms and only 1 equation. I've also tried substituting $$x^2 = 50 – (y+2)^2$$ into the first equation, but the result is

EDIT: I got: $$y^2 + 2y -35 = 0$$ which yields the correct answer. I just made an arithmetic error. Thanks!

I also looked up how to solve for the intersection of circles because that's what these equations remind me of, but this reference (https://mathworld.wolfram.com/Circle-CircleIntersection.html) assumes that one of them is centred at $$(0, 0)$$. I had an idea to substitute $$u = y+2$$, but that just moved the linear terms to another spot in the problem.

Any hints are appreciated to nudge me in the right direction. Thanks in advance!

We have $$x^2-12x+36+y^2=x^2+y^2+4y+4$$ $$3x+y=8$$ Now plug in $$y=8-3x$$ in one of the equations and solve for $$x$$.