Identify the equation of the circle that has its center at (-27,120)and passes through the orgin

The equation of a circle is (x-h)^2 + (y-k)^2 = r^2. Where "x" and "y" are variables, "h" and "k" are the coordinates of the center of the circle, and "r" is the length of the radius. It is given that the center of the circle is (-27, 120). So, h= -27 and k= 120. If the circle passes through the origin, we can assume that the origin is on the circle. Since a circle's radius is constant no matter where it is drawn/is, we can find the radius of the circle by finding the distance between the circle's center (-27, 120) and the origin, (0, 0). The distance formula is: d= √((x[2]-x[1])^2-(y[2]-y[1])^2). If the coordinates of the center of the circle are (x[2}, y[2]), then x[2]= -27 and y[2]= 120. Then, the origin is the (x[1], y[1]). So, x[1] = 0 and y[1] = 0. Plugging the numbers in we get: √((-27-0)^2-(120-0)^2). This gives us √(729+14400) = 123. So since the distance between the center of the circle and a point on the circle is 123 (units), then the radius has a value of 123.

Plugging all the numbers into the equation of a circle, we get: (x-(-27))^2+(y-120)^2=123^2.