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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!

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    $\begingroup$ Translate the equations so that one of them is centered at the origin, then when you have done what you got to do, translate the answer back to where it started from. As we learned in our introduction to pure maths M203 at the O.U. back in 2002: "bodge it to something you know, solve that thing you know how to solve, then bodge it back again." $\endgroup$ Dec 2, 2021 at 22:43

4 Answers 4

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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 $x$.

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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.)

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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:

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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.]

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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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Just set them equal to eachother:

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

Expand, and solve for $x$

$$(x^2 - 12x+36) + y^2 = x^2 + (y^2 + 4y + 4)$$

Combine like terms

$$x^2 - x^2 - 12x = y^2 - y^2 + 4y + 4 - 36$$

$$- 12x = 4y + -32$$

$$x = -\frac{y}{3} + \frac{8}{3}$$

Sub $x$ into a previous equation, solve for $y$, and from the first equation you can then solve for $x$

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In this particular case you can notice that $a^2+b^2=50$ has integer solutions (just test all integers $<\sqrt{50}$ so up to $7$, this is quickly done) namely $$(\pm 1,\pm 7)\text{ and }(\pm 5,\pm 5)$$

Two circles have at most two intersection points, so you can try if the values found above fit the two equations simultaneously and it happens that some sign combinations do, and you don't have to search further.

It is the same as finding obvious roots first when trying to factorize polynomials, then if nothing was found go for calculation.

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