SOLUTION: I am suppose to find the center (h,k) and the radius of the circle given by the equation: x^2-6x+y^2+10y=-30 by completing the square.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I am suppose to find the center (h,k) and the radius of the circle given by the equation: x^2-6x+y^2+10y=-30 by completing the square.      Log On


   

Question 187927: I am suppose to find the center (h,k) and the radius of the circle given by the equation: x^2-6x+y^2+10y=-30 by completing the square.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Find the center and the radius of the circle given by the equation:
x%5E2-6x%2By%5E2%2B10y+=+-30 First group the x- and y-variables as shown:
%28x%5E2-6x%29%2B%28y%5E2%2B10y%29+=+-30 Now complete the squares in both the x- and y-groups by adding the square of half the x-coefficient and the square of half the y-coefficient to both sides of the equation.
Simplify this.
%28x%5E2-6x%2B9%29%2B%28y%5E2%2B10y%2B25%29+=+-30%2B9%2B25 Factor the x-trinomial and factor the y-trinomial and simplify the right side.
%28x-3%29%5E2+%2B+%28y%2B5%29%5E2+=+4 Now compare this with the standard form of the equation of a circle with its center at (h, k) and whose radius is r.
%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2 and you can readily see that:
h+=+3
k+=+-5
r+=+2
So the center of the circle is at (3, -5) and the radius is 2.