SOLUTION: Tutor could you help me please my problem is Find the center and the radius of the circle x^2 +2x+y^2-6y-6=0 I am lost on this one I started with g=1 F=-3 (I think..not sure

Algebra ->  Linear-equations -> SOLUTION: Tutor could you help me please my problem is Find the center and the radius of the circle x^2 +2x+y^2-6y-6=0 I am lost on this one I started with g=1 F=-3 (I think..not sure      Log On


   

Question 135917: Tutor could you help me please my problem is Find the center and the radius of the circle x^2 +2x+y^2-6y-6=0
I am lost on this one I started with g=1 F=-3 (I think..not sure
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B2x%2By%5E2-6y-6=0

Step 1: Put the x terms together and the y terms together. This is already done.

Step 2: Move the constant term to the right side. Add 6 to both sides.
x%5E2+%2B2x%2By%5E2-6y=6

Step 3: Complete the square on the x terms by adding a constant c to make x%5E2%2B2x%2Bc a perfect square.

Step 3a: Divide the coefficient on the x term by 2: 2%2F2=1, then square this value: 1%5E2=1

Step 3b: Add the result of Step 3a to both sides of the equation:
x%5E2+%2B2x%2B1%2By%5E2-6y=6%2B1

Step 4: Repeat step 3 on the y terms:
-6%2F2=-3, -3%5E2=9
x%5E2+%2B2x%2B1%2By%5E2-6y%2B9=6%2B1%2B9

Step 5: Factor the two trinomials:
%28x%2B1%29%5E2%2B%28y-3%29%5E2=16

The general equation of a circle with center at (h, k) and radius r is
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2

So, rewriting our equation slightly: %28x-%28-1%29%29%5E2%2B%28y-3%29%5E2=4%5E2

We can see that the center is at (-1,3) and the radius is 4.