The circle centered at (2,-1) and with radius 4 intersects the circle centered at (2,5) and with radius \((2,-1)\) at two points A and B . Find\((AB)^2\) .

tertre
Apr 15, 2017

#1**0 **

We have the following two equations

(x - 2)^2 + (y + 1)^2 = 16

(x - 2)^2 + (y - 5)^2 = 16

Setting the two equations equal and subtracting (x-2)^2 from both sides. we have that

(y + 1)^2 = (y - 5)^2

y^2 + 2y + 1 = y^2 - 10y + 25 rearrange

12y = 24

y = 2

And subbing this into either of the original equation we have that

(x - 2)^2 + 3^2 = 16

(x - 2)^2 = 7 take both roots

x - 2 = ±√7 add 2 to both sides

x = ±√7 + 2

So....the intersection points are A, B = (√7 + 2, 2) and ( -√7 + 2, 2)

My question, tertre, is how to interpret (AB)^2.......are we supposed to take the dot product of these points and square that???

If so..... [A (dot) B]^2 = [ 4 - 7 + 4]^2 = 1^2 = 1

CPhill
Apr 16, 2017