Question

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

Answer 1

Answer:

  (x +8)^2 +(y -15)^2 = 289

Step-by-step explanation:

The numbers 8, 15, 17 are a Pythagorean Triple, so we know the radius of the circle is 17. Filling in the given information in the standard equation of a circle, we get ...

  (x -h)^2 +(y -k)^2 = r^2 . . . . . . circle with center (h, k) and radius r

  (x +8)^2 +(y -15)^2 = 289 . . . . . circle with center (-8, 15) and radius 17

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Once you have identified the center (h, k)=(-8, 15) and a point you want the circle to go through (x, y)=(0, 0), evaluate the equation for the circle to find the square of the radius:

  (0 +8)^2 +(0 -15)^2 = r^2 = 64+225 = 289