Question

What is the equation in standard form for the circle with center (0, 0) that goes through the point (-5, 12)?

Answer 1

Start with the standard equation of a circle with center at (h,k) and radius r:

(x-h)^2 + (y-k)^2 = r^2. Substitute -5 for h and 12 for k:

(x-0)^2 + (y-0)^2 = r^2

Next, we substitute -5 for x and 12 for y, with the objective of determining the value of r^2:

(-5-0)^2 + (12-0)^2 = r^2. Then 25 + 144 = 169 = r^2, and r = 13.

Thus, the answer choice "x2 + y2 = 169" is the correct one.

Important: Please use " ^ " to denote exponentiation: x^2 + y^2 = 169.