The circle centered at $(2,-1)$ and with radius $4$ intersects the circle centered at $(2,5)$ and with radius $\sqrt{10}$ at two points $A$ and $B$. Find $(AB)^2$.
+128460 Equation of first circle :
(x - 2)^2 + ( y + 1)^2 = 16
Equation of second circle :
(x - 2)^2 + (y - 5)^2 = 10
Subtract the second equation from the first and we have
(y + 1)^2 - ( y - 5)^2 = 6 simplify
y^2 + 2y + 1 - y^2 + 10y - 25 = 6
12y - 24 = 6
12y = 30
y = 30/12 = 2.5
Using the first equation to find the x intersection points, we have that
(x - 2)^2 + ( 2.5 + 1)^2 = 16
(x - 2)^2 + (3.5)^2 =16
(x - 2)^2 + 12.25 =16
(x -2)^2 = 3.75
(x - 2)^2 = 15/4 take both roots
x - 2 = ±√15/ 2
x = ±√15/2 + 4/2
x = [ 4 +√15] / 2 and x = [ 4 - √15] / 2
The distance between these points = [ 4 + √15 ] / 2 - [ 4 - √15] / 2 = 2√15/ 2 = √15
So.....the square of this distance = 15
