SOLUTION: i am trying to find the center and radius of X2+Y2+4X-8Y+13=0 So far i think its (x+2)^2+(y-4)^2=-5^2 meaning the center would be (-2,4) and radius -5

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: i am trying to find the center and radius of X2+Y2+4X-8Y+13=0 So far i think its (x+2)^2+(y-4)^2=-5^2 meaning the center would be (-2,4) and radius -5      Log On


   

Question 643602: i am trying to find the center and radius of X2+Y2+4X-8Y+13=0
So far i think its (x+2)^2+(y-4)^2=-5^2 meaning the center would be (-2,4) and radius -5
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
You have the center correct but not the radius.

x%5E2%2By%5E2%2B4x-8y%2B13=0
We start by completing the squares for both the x terms and the y terms. Usually this tarts with "moving" the constant term to the other side. Subtracting 13 from each side we get:
x%5E2%2By%5E2%2B4x-8y+=+-13

Then we rearrange the terms so that the x terms are together and the y terms are together:
x%5E2%2B4x%2By%5E2-8y+=+-13

To complete the square for the x terms we take half the coefficient of x, 4, and square it. Half of 4 is 2 and 2 squared is, coincidentally, 4. We do the same for the y term. Half of -8 is -4 and -4 squared is 16. So we will add a 4 and a 16 to each side:
x%5E2%2B4x%2B4y%5E2-8y+%2B16=+-13%2B4%2B16
Next we simplify the right side and rewrite the left side as binomial squares:
%28x%2B2%29%5E2%2B%28y-4%29%5E2+=+7
Last of all we rewrite the right side as a perfect square:
%28x%2B2%29%5E2%2B%28y-4%29%5E2+=+%28sqrt%287%29%29%5E2

So the center is (-2, 4) but the radius is sqrt%287%29.