SOLUTION: Write an equation of the circle whose center is at the origin and the point (3,-6) is on the circle. I know the the formula to use is this: (x-h)squared + (y-k)squared = r sq

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Write an equation of the circle whose center is at the origin and the point (3,-6) is on the circle. I know the the formula to use is this: (x-h)squared + (y-k)squared = r sq      Log On


   

Question 39797: Write an equation of the circle whose center is at the origin and the point (3,-6) is on the circle.
I know the the formula to use is this:
(x-h)squared + (y-k)squared = r squared
Found 2 solutions by Nate, fractalier:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
The radius is the distance from the middle to one point on the circle, so use the formula for distance:
d+=+sqrt%28%28y+-+y%29%5E2+%2B+%28x+-+x%29%5E2%29
d+=+sqrt%28%28-6+-+0%29%5E2+%2B+%283+-+0%29%5E2%29
d+=+sqrt%2845%29
The formula for the circle: y%5E2+%2B+x%5E2+=+45

Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Yes the equatin is correct. But the point (h, k) is the center of the circle. Here that is (0. 0).
But since the point (3, -6) lies on the circle, its distance from the origin is the radius of the circle.
Its distance is found by
d^2 = 3^2 + (-6)^2
d^2 = 9 + 36
d^2 = 45
and since d^2 is really r^2, we have
x^2 + y^2 = 45