Give the equation of the circle centered at the origin and passing through the point (0,-8)

Answer 1

Hi Laura H.,

The standard form equation of a circle is (x - h)² + (y - k)² = r², centered at (h, k) with radius r.

This circle is centered at the origin (0, 0), therefore (h, k) = (0, 0) and the equation becomes x² + y² = r².

To find r² we can use the point given (0, -8) for (x, y) and solve for r².

x² + y² = r²

0² + (-8)² = r²

64 = r²

We can now put the equation into its complete form:

x² + y² = 64, the equation of a circle centered at (0, 0) and passing through point (0, -8).

I hope this helps, Joe.

Answer 2

Hi Laura,

If the center of the circle is at the origin (0, 0) and it passes through (0, -8) this means that the radius of the circle is 8. Because the radius is the distance from the center to any point on the circle. The distance between the points (0, 0) and (0, -8) is 8.

The equation of a circle with center (a,b) and radius R is

(x-a)^2 + (y-b)*2 = R2

We have a=b=0 and R=8

so the answer is x^2 + y^2 = 64