[murcorm]

Fine the Center and Radius of the Circle given that it equation is

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

[ebaines]

Your objective is to turn the equation that you were given into an equation of the form:

(x-a)^2 + (y-b)^2 = r^2.

In other words:

x^2 -2ax + a^2 +y^2 - 2by +b^2 = r^2

Once you have this, the center of the circle is at (a,b) and the radius is r.

Let us know how it goes.

[iamthetman]

Well, the standard form for the equation of a circle is x^2 + y^2 = r^2 where the centre of the circle is (0,0) so putting your equation into this format would give you the answer to both your questions.

Do you remember completing the square?
Let's just look at the x's for now:
x^2 + 2x
= (x+1)(x+1) - 1
= (x+1)^2 - 1
This means that the x-coordinate of the centre of the circle is -1 since (x+1)=0 implies x=-1.

Now look at the y's:
y^2 - 6y
= (y-3)(y-3) - 9
= (y-3)^2 - 9
This means that the y-coordinate of the centre of the circle is 3 since (y-3)=0 implies y=3.

Now putting everything together we have:
x^2 + y^2 + 2x -6y - 6 = 0
x^2 + 2x + y^2 - 6y - 6 = 0
(x+1)^2 - 1 + (y-3)^2 - 9 - 6 = 0
Grouping all the like terms gives us:
(x+1)^2 + (y-3)^2 = 16

Now we have standard form, so we know our radius is 4!