Standard?????

Question

Determine the standard form of the equation of a circle with a center at (4,-1) and passing through (0,2).

So I have (x-4)^2 + (y+1)^2 = ?
I don't know what to put for the ? or how to figure it out.

Answer 1

What you put there is the square of the radius of the circle. The radius is simply the distance between the center and any point on the circle - i.e. the distance between (4, -1) and (0, 2). By the pythagorean theorem, this is √((4-0)²+(-1-2)²), and so the square of the radius is (4-0)²+(-1-2)², which is 25.

So the standard form is (x-4)² + (y+1)² = 25.

Answer 2

The equation of a circle is: (x-h)^2+(y-k)^2=r^2

You did the center right. But now that you have the center, you have to figure out the distance from the center to the point on the circle, which would be your radius.

To get the distance, we do the square root of ( (4-0)^2 + (-1-2)^2), which is the square root of 25, or 5.

Your radius is 5, so your equation should look like:

(x-4)^2 + (y+1)^2 =25

Answer 3

To find the radius, use the distance formula from the center to any point on the circle.

r = sqrt[(4-0)^2 + (-1-2)^2] = sqrt(16 + 9) = sqrt 25 = 5

In the eqn. for a circle : (x-h)^2 + (y-k)^2 = r^2

So (x-4)^2 + (y+1)^2 = 25