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The points (-3,2) and (-2,3) lie on a circle whose center is on the x-axis. What is the radius of the circle?

 

Pls explain.

 Dec 22, 2018
edited by Guest  Dec 22, 2018
 #1
avatar+100473 
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Lot the radius be located at (a, 0)

 

Since the distance from the center to each of these points is the same, we can use the equaion of a circle to find a

 

(-3 - a)^2  + ( 0 - 2^2 = r^2

(-2 - a)  + ( 0 - 3)^2  = r^2

 

Since r^2 = r^2.....then

(-3 - a)^2 + (0 -2)^2  =  (-2-a)^2 + (0 -  3)^2       simplify 

 

a^2 + 6a + 9  + 4  =  a^2 + 4a + 4  + 9      subtract like terms from each side

 

6a   = 4a

6a - 4a = 0

a(6-4) = 0

2a = 0

a = 0

 

So...the center is (0,0)

 

And  the radius is

 

sqrt [ (-2 - 0)^2 + (0 - 3)^2 ]   =  sqrt [ 2^2 + 3^2]  = sqrt (13)

 

 

cool cool cool

 Dec 22, 2018
 #2
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thanks for the answer. :)

 Dec 22, 2018

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