SOLUTION: What is the center and the radius of a circle with the equation: x^2 + 6x + y^2 - 8y - 11 = 0

Algebra ->  Coordinate-system -> SOLUTION: What is the center and the radius of a circle with the equation: x^2 + 6x + y^2 - 8y - 11 = 0      Log On


   



Question 10828: What is the center and the radius of a circle with the equation:
x^2 + 6x + y^2 - 8y - 11 = 0

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The goal here is to get the equation of a circle in the standard form:
%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2 This is the equation of a circle with radius r and center at (h, k).
The technique is the "complete the square" in x and y.
x%5E2+%2B+6x+%2B+y%5E2+-+8y+-+11+=+0 Isolate the x and y terms by adding 11 to both sides.
%28x%5E2+%2B+6x%29+%2B+%28y%5E2+-+8y%29+=+11 Now complete the squares in x and y. This process entails adding a constant term to the x-group and to the y-group so that when you factor the x-group and the y-group, you will have a binomial squared in x and a binomial squared in y.
The constant for the x-group is found by squaring one-half of the coefficient of the x-term. That's (6/2)^2 = 9. Similarly for the y-group, you square one-half of the coefficient of the y-term. That's (-8/2)^2 = 16.
%28x%5E2+%2B+6x+%2B+9%29+%2B+%28y%5E2+-+8y+%2B+16%29+=+11+%2B+9+%2B+16 Don't forget to add the same numbers to both sides of the equation.
%28x%5E2+%2B+6x+%2B+9%29+%2B+%28y%5E2+-+8y+%2B+16%29+=+36 Now factor.
%28x+%2B+3%29%5E2+%2B+%28y+-+4%29%5E2+=+6%5E2 Now it looks like the standard form of the equation for a circle with radius of 6 and center at (-3, 4).