<s>[INST] How to solve this system of equations by hand: $(x-6)^2 + y^2 = 50, x^2 + (y+2)^2 = 50$ 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!
 [/INST] Since the circles have the same radius, their intersection points will lie on the perpendicular bisector of the segment connecting the centers of the circles.  That midpoint is $ \ \left(\frac{6 \ + \ 0}{2} \ , \ \frac{0 \ + \ [-2]}{2} \right) \ = \ (3 \ , \ -1) \ \ . $  The slope of the line segment is $ \ \frac{0 \ - \ [-2]}{6 \ - \ 0} \ = \ \frac13 \ \ , $ so the slope of the perpendicular bisector is $ \ -3 \ \ , $ making its equation
$$ y \ - \ [-1] \  = \  -3·(x \ - \ 3) \ \ \Rightarrow \ \ y \ = \ -3x \ + \ 8 \ \ . $$
Inserting this result into either circle equation produces
$$ ( x - 6)^2 \ + \ (-3x + 8 )^2 \ = \ 50 \ \ or \ \ x^2 \ + \ (-3x + 10 )^2 \ = \ 50 $$ $$\Rightarrow \ \ 10x^2 \ - \ 60x \ + \ 50 \ = \ 0 \ \ \rightarrow \ \ x^2 \ - \ 6x \ + \ 5 \ = \ 0 \ \ . $$
This factors pretty easily to yield the $ \ x-$coordinates for the intersections; using those in the bisector line equation then gives us the $ \ y-$ coordinates.  (Both intersection points have integer coordinates.)

$ \ \ $
ADDENDUM (12/5) --
What if the two radii are unequal?  The intersection points of the two circles still lie on a line perpendicular to the segment connecting their centers, but that line does not bisect the segment.  If we call the larger circle's radius $ \ R \ \ , $  the smaller radius $ \ r \ \ , $ and the distance between the centers $ \ D \ \ , $ then the triangle formed by those two centers and one of the intersection points resembles the following:

In this diagram,  $ \ h \ $ is the distance of an intersection point from the center-linking segment; we wish to determine the point where the intersection-linking line meets this segment, with $ \ x \ $ being the distance of that point from the center of the smaller circle.  Because the triangle is divided into two smaller right triangles, we can write
$$ h^2 \ \ = \ \ r^2 \ - \ x^2 \ \ = \ \ R^2 \ - \ (D - x)^2 \ \ \Rightarrow \ \  r^2 \ - \ x^2 \ \ = \ \ R^2 \ - \ D^2 \ + \ 2Dx \ - \ x^2  $$
$$ \Rightarrow \ \  x  \ \ = \ \ \frac{D^2 \ - \ (R^2 \ - \ r^2)}{2D} \ \ \ \text{or} \ \ \ \frac{x}{D} \ \ = \ \ \frac{D^2 \ - \ (R^2 \ - \ r^2)}{2D^2} \ \ . $$
[We note that this confirms the earlier remark that the line of intersections bisects the center-linking segment when the two circles are of equal radius.]

As an illustration for this method, we find the intersections of the two circles $ \ (x-4)^2 + y^2 \ = \ 36 \ \ $ and $ \ x^2 + (y+3)^2 \ = \ 25 \ \ . $  As the centers of the circles are $ \ (4 \ , \ 0 ) \ $ and $ \ (0 \ , \ -3) \ \ , $ the distance separating them is $ \ D = 5 \ \ . $  The relation we derived gives us
$$  \frac{x}{D} \ \ = \ \ \frac{25 \ - \ (36 \ - \ 25)}{2 \ · \ 25} \ \ = \ \ \frac{25 \ - \ 11}{50} \ \ = \  \ \frac{7}{25} \ \ . $$
So the intersection of the two lines of interest is $ \ \frac{7}{25} \ $ of the way from $ \ (0 \ , \ -3) \ $ to $ \ (4 \ , \ 0) \ \ $ or $ \ \left(\frac{18·0 \ + \ 7·4}{25} \ , \ \frac{18·[-3] \ + \ 7·0}{25} \right) \ = \ \left(\frac{28}{25} \ = \ 1.12 \ , \ -\frac{54}{25} \ = \ -2.16 \right) \ \ . $  The slope of the segment connecting the circle centers is $ \ \frac{0 \ - \ [-3]}{4 \ - \ 0} \ = \ \frac34 \ \ . $ Hence, the slope of the line of intersections is $ \ -\frac43 \ \   $ and its equation is
$$ y \ - \ \left[-\frac{54}{25} \right] \  = \  -\frac43·\left(x \ - \ \frac{28}{25} \right) \ \ \Rightarrow \ \ y \ = \ -\frac43x \ - \ \frac23 \ \ . $$
Upon inserting this relation into, say, the first line equation yields
$$ (x \ - \ 4)^2  \ + \ \left(-\frac43x \ - \ \frac23 \right)^2 \ = \ 36 \ \ \Rightarrow \ \ 25x^2 \ - \ 56x \ - \ 176 \ \ = \ \ 0 $$ $$ \Rightarrow \ \ x \ = \ \frac{28 \ \pm \ 72}{25} \ \ . $$
Using the equation of the line of intersections, we find the coordinates of the intersection points to be
$$ (4 \ , \ -6 ) \ \ \ \text{and} \ \ \ \left(-\frac{44}{25} \ = \ -1.76 \ , \ \frac{42}{25} \ = \ 1.68   \right) \ \ .  $$  The graph above presents the geometrical situation.
[This is not presented as a more efficient method of solving for the circle intersections, but as an alternative way of viewing the problem.]
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