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?
(-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