Question

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 $(AB)^2$.

Answer 1

The first circle has equation

(x - 2)² + (y + 1)² = 4²

and the second has equation

(x - 2)² + (y - 5)² = (√10)²

Solve for (x - 2)² :

(x - 2)² + (y + 1)² = 4² ==> (x - 2)² = 16 - (y + 1)²

(x - 2)² + (y - 5)² = (√10)² ==> (x - 2)² = 10 - (y - 5)²

Then

16 - (y + 1)² = 10 - (y - 5)²

16 - (y ² + 2y + 1) = 10 - (y ² - 10y + 25)

15 - 2y - y ² = -15 + 10y - y ²

30 - 12y = 0

12y = 30

y = 30/12 = 5/2

(this is the y coordinate of A and B)

Then solve for x :

(x - 2)² = 16 - (5/2 + 1)²

(x - 2)² = 15/4

x - 2 = ± √(15/4) = ±√15/2

x = 2 ± √15/2

(these are the x coordinates for either A or B)

The intersections are the points A = (2 - √15/2, 5/2) and B = (2 + √15/2, 5/2). We want to find the squared distance between them:

(AB)² = [(2 - √15/2) - (2 + √15/2)]² + (5/2 - 5/2)²

(AB)² = (-√15)² + 0²

(AB)² = 15