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.

Guest Dec 22, 2018

edited by
Guest
Dec 22, 2018

#1**+1 **

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)

CPhill Dec 22, 2018