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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?

Dec 23, 2020

#1
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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?

r = √13

Dec 23, 2020
#2
+114318
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If the center lies on the x axis, call the   center  (a, 0)

So.....the distance from this point to each of the given points  is equal

So....equating distances, we have that

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

((-1) (3 + a))^2  + 2^2  =  ((-1)(2 + a))^2  + 3^2

(3 + a)^2  + 4  = (2 + a)^2  +  9

9+ 6a + a^2  + 4  = 4 + 4a + a^2 + 9        subtract a^2  from both sides and simplify

6a +13 = 4a + 13     subtract 13 from both sides

6a  = 4a

This is true when a  = 0

So  the center is    (a, 0)   = (0, 0)

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

Here's a graph :   https://www.desmos.com/calculator/dd6nbqzagb

Dec 23, 2020
#3
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Thank you so much.

Nice explanation too.

Spicy3.14  Dec 23, 2020
#4
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r = sqrt(-32 + 22)     or    r = sqrt(-22 + 32)  ==>  r = sqrt(13)

Dec 23, 2020