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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 $(AB)^2$.

 Apr 12, 2021
 #1
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Equation of circle centered at (h,k) with radius r:

\[(x - h)^2 + (y - k)^2 = r^2.\]

 

$(AB)^2 = 18$

 Apr 13, 2021
 #2
avatar+1641 
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The circle centered at (2, -1) and with radius 4 intersects the circle centered at (2, 5) and with radius √10 at two points A and B. Find (AB)2.

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Let the centers of the circles be   P(2, -1) and Q(2, 5)       PQ = |-1 + 5| = 6

 

Let the midpoint of AB be M.     AM = BM ==> x

 

sqrt(AP2 - x2) + sqrt(AQ2 - x2) = PQ              x = (√15) / 2

 

AB2 = (2x)2 

 

AB2 = 15

 Apr 13, 2021

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