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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
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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

 

 

 

cool cool cool

CPhill  Apr 16, 2017

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