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The two circles are:

1) $$(x-2)^2 + (y+1)^2 = 25$$ 2) $$(y-2)^2 + (x+1)^2 = 25$$


marked as duplicate by user228113, Claude Leibovici, Frunobulax, Harish Chandra Rajpoot, Pierre-Guy Plamondon Feb 22 '16 at 12:40

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  • $\begingroup$ Set $(1) = (2)$. You will get a relation between $x$ and $y$. Substitute in any equation to get the point(s) of intersection. $\endgroup$ – SS_C4 Feb 22 '16 at 8:01
  • $\begingroup$ Or you could just graph the circles and check. $\endgroup$ – SS_C4 Feb 22 '16 at 8:06
  • $\begingroup$ @Antonios-AlexandrosRobotis We can apply a trick by flipping the role of the axes but the mentioned cannot. $\endgroup$ – Ng Chung Tak Feb 22 '16 at 8:33
  • $\begingroup$ Perhaps I've misunderstood your comment, but the answer I linked should solve this question in generality: regardless of tricks or not. In either case, I do believe it answers this question. $\endgroup$ – Antonios-Alexandros Robotis Feb 22 '16 at 8:35

$(1)$ and $(2)$ are symmetric for $x=y$. The two points you are looking for are the intersenction between $(1)$ and $y=x$.

Therefore, the intersections are $(\frac{1 \pm \sqrt{41}}{2},\frac{1 \pm \sqrt{41}}{2})$.


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