Circle $C_1$ is centered at $(2,-1)$ with radius $4$. Circle $C_2$ is centered at $(-2,5)$ with radius $10$. Circles $C_1$ and $C_2$ intersect at two points $A$ and $B$. Find $(AB)^2$.
+129872 Equation of first circle
(x - 2)^2 + ( y + 1)^2 = 16 → x^2 - 4x + 4 + y^2 + 2y + 1 = 16
Equation of second circle
(x + 2)^2 + (y - 5)^2 = 100 → x^2 + 4x + 4 + y^2 - 10y + 25 = 100
Subtract the second equation from the first
-8x + 12y -24 = -84
-8x + 12y = - 60
-2x + 3y = -15
3y = 2x - 15
y= [ (2/3)x - 5 ] sub this into the first equation for y
(x - 2)^2 + [ (2/3)x - 5 + 1]^2 = 16
(x - 2)^2 + [ (2/3)x - 4]^2 = 16
x^2 - 4x + 4 + (4/9)x^2 - (16/3)x + 16 = 16
(13/9)x^2 -(28/3)x + 4 = 0
13x^2 - 84x + 36 = 0
(13x -6) ( x - 6) = 0
Setting both factors to 0 and solving for x we have that
x = 6/13 and y = (2/3)(6/13) - 5 = 12/39 - 5 = -61/13
x= 6 and y= (2/3)(6) - 5 = -1
(AB)^2 = ( 6 - 6/13)^2 + ( -61/13 + 1)^2 = (72/13)^2 + ( -48/13)^2 = 576 / 13
