Center & Radius of Circle

[lewch45]

x^2 + y^2 + 4x-2y-11=0

I combined like terms:

(x^2+4x) + (y^2-2y)=11

I wasn't sure where to go from here. An example says to complete the square but I'm not sure how to do that.

 

[Guest]

What are you trying to do, graph it?

 

[stapel]

The general process for completing the square for circles is as follows:

. . . . .1) Group the x-terms together, group the y-terms together,
. . . . .and put the loose numbers on the other side.

. . . . .2) If the squared terms are multiplied by anything,
. . . . .divide the entire equation by this value.

. . . . .3) For the x-terms, take half of the coefficient on "x",
. . . . .and square it. Add to both sides.

. . . . .4) For the y-terms, take half of the coefficient on "y",
. . . . .and square it. Add to both sides.

. . . . .5) Convert the x- and y-terms to squared-binomial
. . . . .form; simplify on the other side. You may also want to
. . . . .convert the numerical side to squared form.

So, for instance:

. . . . .x² + y² - 6x + 8y = 0

. . . . .1) x² - 6x + y² + 8y = 0

. . . . .2) (not needed, in this case)

. . . . .3) (1/2)(-6) = -3, (-3)² = 9
. . . . .x² - 6x + 9 + y² + 8y = 0 + 9

. . . . .4) (1/2)(8) = 4, (4)² = 16
. . . . .x² - 6x + 9 + y² + 8y + 16 = 0 + 9 + 16

. . . . .5) (x - 3)² + (y + 4)² = 25 = 5²

In my example, then, the center is at (h, k) = (3, -4), and the radius r = 5.

Eliz.

 

[Guest]

stapel? Is that you? What are you doing here! I was going to answer that (as soon as I figured out how )

 

[stapel]

stapel? Is that you?

Um... yes, "stapel" is "stapel". Who else did you think "stapel" would be? :shock:

What are you doing here!

Tutoring. It's what tutors do on a tutoring forum. What are you doing here? :roll:

Eliz.

 

[tkhunny]

I combined like terms:

(x^2+4x) + (y^2-2y)=11

No you didn't. You just grouped a few things.