Question

Identify the equation of the circle that has its center at (-16, 30) and passes through the origin.

Answer 1

The general equation of the circle is:

(x -xo)^2 + (y - yo)^2 = r^2

Where xo,yo are the coordinates of the center and r is the radius of the circle.

Here (xo,yo) is (-16,30) and you can find the radius.

You can find the radius using the equation of the distance (Pythagoras) between the center (-16,30) and any point on the circunference. Here use (0,0)

r^2 = (-16 - 0)^2 + (30 - 0)^2 = 1156

Then, the equation is>

(x - (-16) )^2 + (y - 30)^2 = 1156

=> (x + 16)^2 + (y - 30)^2 = 1156.   <----- answer