-5

The two circles are:

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

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

Some Guy
  • 2,728
badatmath
  • 121
  • Set $(1) = (2)$. You will get a relation between $x$ and $y$. Substitute in any equation to get the point(s) of intersection. – SS_C4 Feb 22 '16 at 08:01
  • Or you could just graph the circles and check. – SS_C4 Feb 22 '16 at 08:06
  • @Antonios-AlexandrosRobotis We can apply a trick by flipping the role of the axes but the mentioned cannot. – Ng Chung Tak Feb 22 '16 at 08:33
  • 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. – Alekos Robotis Feb 22 '16 at 08:35

1 Answers1

2

$(1)$ 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 $(\frac{1 + \sqrt{41}}{2},\frac{1 + \sqrt{41}}{2})$ and $(\frac{1 - \sqrt{41}}{2},\frac{1 - \sqrt{41}}{2})$.

Yash
  • 103
S.C.B.
  • 23,106