Question

PLEASE ANSWER QUICK!!! Identify the equation of the circle that has its center at (-8, 15) and passes through the origin.

Answer 1

(X+8)^2+(y-15)^2=289 would be the equation to the circle

Answer 2

(x + 8)^2 + (y - 15)^2 = 17^2

Equation of a circle
Midpoint at (a,b)

(x - a)^2 + (y - b)^2 = r^2

You know what the circle’s center is so you can substitute those values in

(x + 8)^2 + (y - 15)^2 = r^2

But you don’t know what the radius is. To find the radius, we know that the circle passes through the origin (0,0) so you just use the distance formula to figure out the radius length.

Distance formula:

Square root of (x1 - x2)^2 + (y1 - y2)^2

So...

Square root of (-8 - 0)^2 + (15 - 0)^2
Square root of 64 + 225
Square root of 289
r=17

So you substitute 17 into the equation of a circle formula and the result is:

(x + 8)^2 + (y - 15)^2 = 17^2

~~hope this helps~~