precalculus help pls

Question

Write the standard form of the equation of the circle with the given characteristics.

Center: (0, -1); Solution point (3,2)

Answer 1

Hello Morgan!

remember that the circle has the properties that all points in the circle are in the same distance from the center, which is the radius of the circle. Let's call the radius by the letter r

From the equation of the distance between two points, we have:

r² = (3-0)² + (2- (-1))²

r² = 9 + 9 = 18

If all the points in the circle are in the same distance of the center, we have

(x- 0)² + (y - (-1))² = r² = 18

x² + (y+1)² = 18

Answer 2

A circle is x^2 + y^2 = r^2, where r is the radius, and the center is at (0,0)

Since the center is at (0,-1), we need a y-offset, so it looks like this:

x^2 + (y + 1)^2 = r^2

To find r, simply plug in the "solution point" supplied

CHECK by graphing---using a graphing calculator or something like the desmos online tool.