Question

Find the center and radius for the circle, given the equation: x^2-14x+y^2+8=16

Answer 1

To find the center of a circle, we must first convert into standard form, in which: (x-h)^2 + (y-k)^2 = r^2, where the center is at (h, k) and the radius=r
So we need to convert our equation x^2-14x+y^2+8=16 into that format.
First, separate the x-terms from the
y-terms: x^2-14x
y^2-0
+8-8=16-8 --> = 8
Second, how do we find the factors for x^2-14x+c and y^2-0x+c ?
Take half of 14 and square it: -7×-7=49=c --> so x^2-14x+49 = (x-7)^2
Y is easier because we have no
x-term (since it's 0x), so y^2 +/-0
= (y-0)^2 which = y^2
Third, we have to find the constant to make the 49 work. Since originally we got it to = 8, we needed 49 to complete the (x-7) square. So add 49 to the right side of the "=" and you get 49+8=57
Now it looks like: x^2-14x+49+y^2=57, but we know that it becomes:
(x-7)^2 + (y-0)^2 = 57
Now it is in standard circle format, and our center (h, k) is (7, 0), and our radius becomes the square root of 57.