Question

The center of a circle is located at (3, 8), and the circle has a radius that is 5 units long. What is the general form of the equation for the circle? A. x2 + y2 − 6x − 16y + 48 = 0 B. x2 + y2 − 6x − 16y − 25 = 0 C. x2 + y2 + 6x + 16y + 48 = 0 D. x2 + y2 + 6x + 16y − 25 = 0

Answer 1

A. x² +y² -6x -16y +48 = 0

Explanation:

The standard-form equation for a circle centered at (h, k) with radius r is ...

(x -h)² +(y -k)² = r²

For your circle, this is ...

(x -3)² +(y -8)² = 5²

To put this in general form, you subtract the constant on the right, and eliminate parentheses:

x² -6x +9 +y² -16x +64 -25 = 0

x² +y² -6x -16y +48 = 0 . . . . . rearrange to descending powers of x, y