Find the standard form of the circle x^2+y^2+2x-6y-6=0.?

Answer 1

x^2 + y^2 + 2x - 6y - 6 = 0

We're inevitably going to complete the square, so the first thing we need to do is rearrange the terms so that the x terms are together and the y terms are together.

x^2 + 2x + y^2 - 6y - 6 = 0

We need to complete the square with the x terms, and complete the square with the y terms. We look at the coefficient of x (which is 2), and the coefficient of y (which is -6), and add "half squared" of each of those terms. Half squared of 2 is 1, and half squared of -6 is 9, so we add 1 and 9 to both sides.

x^2 + 2x + 1 + y^2 - 6y + 9 - 6 = 1 + 9

Change into binomial squares,

(x + 1)^2 + (y - 3)^2 + - 6 = 10
(x + 1)^2 + (y - 3)^2 = 16

Edit: Thanks for pointing out the correct to the answerers below.

Answer 2

The standard form for the circle is (x - a)^2 + (y - b)^2 = r^2, where the center is (a,b) and the radius is r. You want to put x^2 + y^2 + 2x - 6y - 6 = 0 into this form.

Group the x terms together. Do the same with the y terms. Move the rest to the other side:

x^2 + y^2 + 2x - 6y - 6 = 0
(x^2 + 2x) + (y^2 - 6y) = 6

Now use completing-the-square on each group on the left, so that you can get the first group into the form of (x-a)^2 and the second into the form of (y-b)^2. You should end up with "1" having to be added to the first group, and "9" being added to the second group in order to complete the squares:

(x^2 + 2x + 1 -1) + (y^2 - 6y + 9-9) = 6
(x^2 + 2x + 1) -1 + (y^2 - 6y + 9) -9 = 6
(x+1)^2 - 1 + (y - 3)^2 - 9 = 6
(x+1)^2 + (y - 3)^2 = 16
(x+1)^2 + (y - 3)^2 = 4^2

Answer 3

Complete the squares!

x² + y² + 2x - 6y - 6 = 0
→ (x² + 2x + 1) + (y² - 6y + 9) = 16
→ (x+1)² + (y-3)² = 16.

Radius 4, centered at (-1,3).

Answer 4

Puggy, I don't know how you mis-subtracted after that brilliant answer, but the RHS should equal 16 when all is said & done.