What is the solution of (x-1)^2 + (y+2)^2 = (x+3)^2+ (y-6)^2?

Here we have to solve: (x-1)^2 + (y+2)^2 = (x+3)^2+
(y-6)^2

Let us rearrange the
terms:

=> (x-1)^2 - (x+3)^2 + (y+2)^2 - (y-6)^2 =
0

We use a^2 - b^2 = (a-b)(a+b) and
expand

=> (x-1 - x -3)(x-1 +x +3) + (y+2 -y +6)(y+2
+y -6) = 0

=> -4*( 2x +2 ) + 8 ( 2y -4)
=0

divide by 8

=> -x-1
+ 2y - 4 =0

=> 2y- x -5
=0

Therefore the given expression

(x-1)^2 + (y+2)^2 = (x+3)^2+
(y-6)^2

reduces to 2y- x -5
=0

We cannot solve further and
find the values of x and y