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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?

ant101  Mar 23, 2017
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 #1
avatar+75395 
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(-3,2) and (-2,3)

 

If the center is on the x axis it will have the coordinate  (0, y)

 

And since both points are on the circle, they are equidistant from the center so we can solve this :

 

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

 

9  + 4 - 4y + y^2  =  4 + 9 - 6y + y^2

 

13 -4y + y^2  =  13 - 6y + y^2      subtract  13, y^2 from both sides

 

-4y = -6y     add 6y to both sides

 

6y - 4y  = 0

 

2y  = 0     divide both sides by 2

 

y  = 0

 

So the center is (0,0 )     and using either point, the radius  =

 

sqrt [ ( -3-0)^2 + (2- 0)^2 ]  =

 

sqrt [ 9 +  4]  =  

 

sqrt(13)

 

Here's a graph with the points : 

 

https://www.desmos.com/calculator/in6iwdjaqp

 

 

cool cool cool

CPhill  Mar 23, 2017
 #2
avatar+230 
0

Thanks so much!

ant101  Mar 23, 2017

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