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A circle passes through the points \((-2,0)\)\((2,0)\), and \((3,2)\). Find the center of the circle. Enter your answer as an ordered pair.

Thank you!

May 3, 2020

#1
+21957
+1

A  =  (-2, 0)

B  =  (2, 0)

C  =  (3, 2)

Since there is a right angle at B, AC must be a diameter of the circle.

Therefore, the center of the circle is the midpoint of AC.

The midpoint of AC  =  ( (-2 + 3)/2, (0 + 2)/2 )  =  ( 1/2, 1 )

May 3, 2020
#2
0

That was incorrect. Thank you for trying though. Any idea on what went wrong?

May 3, 2020
#3
+21957
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You're right -- I messed up -- I thought (3,2) was (2,3) -- my fault ...

May 3, 2020
#4
0

It's fine. Do you know what the real answer is though?

Guest May 3, 2020
#5
+112185
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Find the midpoint  of  (-2,0)  and (2,0)   =  (0,0)

The  line perpendicular to this point is just the x axis and has the equation x  = 0

Find the  midpoint  of  (2,0)  and(3,2)   = [ 3+2/2, 2+0/2)  = (5/2, 1)

The slope of the  line  through these points  = [ 2-0]/[3 - 2] =  2

And the slope of a perpendicular line passing through the  midpoint we just found has the slope -1/2

The equation of this line is

y = (-1/2) ( x - 5/2) + 1

Letting x  = 0  we can find the  y coordinate of the  center of the  circle

y = (-1/2)(0 -5/2) + 1  = 5/4 + 1  = 9/4

So...the center of the circle is  (0, 9/4)

And  the square of the radius is  2^2   + (9/4)^2  =  4+ 81/16  = 145/16

Here's the graph  https://www.desmos.com/calculator/wdbs8u2edh

May 3, 2020
edited by CPhill  May 4, 2020
#6
+1

Thank you!

Guest May 3, 2020