# Please find the center and radius of the circle.?

x^2 + y^2 + 8x + 2y +18 = 0

I am very sure I typed the right thing. And this is a published equation from my math book, too.

### 1 Answer

- Anonymous9 years agoFavorite Answer
x^2 + y^2 + 8x + 2y +18 = 0

(x^2 + 8x) + (y^2 + 2y) = -18

(x^2 + 8x + (8/2)^2 ) + (y^2 + 2y + (2/2)^2 ) = -18 + (8/2)^2 + (2/2)^2

You have to complete the square for x and y. Because I added some extra terms on the LHS, I must add the same amount to the RHS.

(x+4)^2 + (y+1)^2 = -18 + (8/2)^2 + (2/2)^2

(x+4)^2 + (y+1)^2 = -18 + 16 + 1

(x+4)^2 + (y+1)^2 = -1

(You can always expand out this equation to see that it equals x^2+y^2+8x+2y+18=0)

Center is at (-4,-1).

However, the RHS should be a positive number, because the radius is the square root of this number. You can't have a radius of sqrt(-1).

A good reference is http://www.purplemath.com/modules/sqrcircle.htm for the "completing the square" step.

-- Since you are sure that you typed in the correct equation (and I am sure that I have done the working right), it must mean that the question is somehow wrong.

In this situation, it is simply not possible to have a radius of sqrt(-1). However, it is also important to know what you are and aren't supposed to get as a reasonable answer. You can apply the recipe I used in solving the problem (or look at the reference) for finding the center and radius of circles.

Why don't you try some other examples from your textbook? See if this method will work for those questions too. Good luck.