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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?

 Nov 24, 2018
 #1
avatar+101424 
+2

Let the center be (x, 0)

 

The points will be equidistant from the center..so....we can solve this

 

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

 

x^2 + 6x + 9 + 4   = x^2 + 4x + 4 + 9    

 

6x = 4x   ⇒   x = 0

 

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

 

And the radius   is     sqrt ( (-3 - 0)^2 + ( 2 - 0)^2 )  =  sqrt [ 9 + 4 ]   = sqrt (13)

 

 

cool cool cool

 Nov 24, 2018
 #2
avatar+814 
+1

Thank you, CPhill! Now I understand better!

mathtoo  Nov 24, 2018
 #3
avatar+101424 
0

OK.......good deal.....!!!

 

 

cool cool cool

CPhill  Nov 24, 2018

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