Question

A circle centered at (-2,3) and contains point (5,9) which represents the equation

Answer 1

The standard form of the equation of a circle is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle, (x,y) is a point of the circle, and r is the length of the radius of the circle. When the equation of a circle is written, h,k, and r are numbers, while x and y are still variables. (x-2)^2 + (y-k)^2 = 16 is an example of a circle. The problem gives us two of the three things that a circle has, a point (5,9) and the center (-2,3). We need to find the radius in order to write the equation. We substitute -2 for h, 3 for k, 5 for x, and 9 for y to get (5 - (-2))^2 + (9 - 3)^2 = r^2 We simplify: 49 + 36 = r^2, r^2 = 85. We only need to know r^2 because the equation of a circle has r^2. We now have all the information to write the equation of a circle. (x + 2)^2 + (y - 3)^2 = 85.