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.
+130071 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)
