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Circle $C_1$ is centered at $(2,-1)$ with radius $4$.  Circle $C_2$ is centered at $(-2,5)$ with radius $10$.  Circles $C_1$ and $C_2$ intersect at two points $A$ and $B$. Find $(AB)^2$.

 Aug 15, 2023
 #1
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Equation of first circle

(x - 2)^2 + ( y + 1)^2 =  16   → x^2 - 4x + 4 + y^2 + 2y + 1  =  16

Equation of second  circle

(x + 2)^2  + (y - 5)^2 = 100 → x^2 + 4x + 4 + y^2 - 10y + 25 = 100

 

Subtract the second equation  from the  first

 

-8x + 12y -24 = -84

 

-8x + 12y  = - 60

 

-2x + 3y = -15

 

3y = 2x - 15

 

y=   [  (2/3)x - 5 ]     sub this into the first equation  for  y

 

(x - 2)^2  + [ (2/3)x - 5 + 1]^2  = 16

 

(x - 2)^2  + [ (2/3)x - 4]^2  = 16

 

x^2 - 4x + 4 + (4/9)x^2 - (16/3)x + 16 = 16

 

(13/9)x^2  -(28/3)x + 4  = 0

 

13x^2 - 84x + 36 = 0 

 

(13x -6) ( x - 6) = 0

 

Setting both factors to 0  and solving  for  x we have that

 

x = 6/13     and y = (2/3)(6/13) - 5  =  12/39 - 5  =  -61/13

 

x= 6    and y=  (2/3)(6) - 5 =  -1

 

(AB)^2 =  ( 6 - 6/13)^2 + ( -61/13 + 1)^2  =  (72/13)^2  + ( -48/13)^2  =  576 / 13

 

cool cool cool

 Aug 15, 2023

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