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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.

 Jul 23, 2022
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We have

(x - 2)^2 + ( y + 1)^2 - 4^2  = (x -2)^2 + ( y - 5)^2  - (sqrt 10)^2

 

(y + 1)^2  - 16 =  (y -5)^2  - 10

 

(y + 1)^2  - (y -5)^2  = 6

 

y^2 + 2y + 1  - ( y^2 - 10y + 25)  = 6

 

12y - 24 = 6

 

12y = 30

 

y = (30/12)  = (5/2)  = 2.5

 

And

(x -2)^2  + (y + 1)^2 = 16

 

(x -2)^2  + ( 2.5 + 1)^2  = 16

 

( x - 2)^2  + 12.25   = 16

 

(x -2)^2  = 16 - 12.25

 

(x - 2)^2  = 3.75

 

x - 2 =  sqrt (3.75)             x -2  = -sqrt (3.75)

 

x =  2 + sqrt (3.75)  =  x coord of A          x = 2 - sqrt (3.75)  =  x coord of   B

 

AB =   ( 2 + sqrt (3.75) - ( 2 -sqrt (3.75) ) =  (2sqrt (3.75))

 

(AB)^2 =  (2sqrt (3.75))^2 =   4 (3.75)  =   15

 

 

cool cool cool

 Jul 23, 2022

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