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.
+128474 We have
(x - 2)^2 + ( y + 1)^2 - 4^2 = (x -2)^2 + ( y - 5)^2 - (sqrt 10)^2
(y + 1)^2 - 16 = (y -5)^2 - 10
(y + 1)^2 - (y -5)^2 = 6
y^2 + 2y + 1 - ( y^2 - 10y + 25) = 6
12y - 24 = 6
12y = 30
y = (30/12) = (5/2) = 2.5
And
(x -2)^2 + (y + 1)^2 = 16
(x -2)^2 + ( 2.5 + 1)^2 = 16
( x - 2)^2 + 12.25 = 16
(x -2)^2 = 16 - 12.25
(x - 2)^2 = 3.75
x - 2 = sqrt (3.75) x -2 = -sqrt (3.75)
x = 2 + sqrt (3.75) = x coord of A x = 2 - sqrt (3.75) = x coord of B
AB = ( 2 + sqrt (3.75) - ( 2 -sqrt (3.75) ) = (2sqrt (3.75))
(AB)^2 = (2sqrt (3.75))^2 = 4 (3.75) = 15
