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

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

Let us expand brackets:

==> x^2 -2x + 1 + y^2 + 4y + 4 = x^2 +6x +9+y^2 -12y + 36

Let us reduce similars:

==> -2x + 4y + 5 = 6x -12y + 45

Now combine like terms:

==> 10x +16y = 40

Now divide by 2:

**==> 5x + 8y = 20**

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 **

To find the solution (x-1)^2 + (y+2)^2 = (x+3)^2+ (y-6)^2?

Solution:

We expand both sides:

x^2-2x+1 +y^2+4y+4 = x^2+6x+9+y^2-12y+36.

We collect the liketerms on the left.

(x^2-x^2) +(-2x-6x)+(y^2-y^2) +(4y+12y) +5-(9+36) = 0

-8x +16y-40 = 0

Divide by -8:

x-2y-5 = 0 is the solution for the locus of the intersecting points of two circles (x-1)^2+(y+2)^2 = r^2 and (x-3)^3+(y+6)^2 = r^2 , with fixed centres at (1 ,-2) and (3,-6) respectively and equal radii. The solution is actually the radical axis of the the two family of circles