Equations of Circles

1. Oct 21, 2010

morrowcosom

1. The problem statement, all variables and given/known data
Here is the equation of a circle:
x^2 + y^2 - 3x + 7y - 6 = 0

It is your job to find the center and the radius of the circle.

--------------------------------------------------------------------------------

Enter the coordinates of the center and the radius (as numbers):

2. Relevant equations

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

3. The attempt at a solution
The problem: x^2 + y^2 - 3x + 7y - 6 = 0
(x-3/2)^2+(y-(-7/2)^2

(x-3/2) (x-3/2)= x^2 -3/2x -3/2x +9/4= x^2-3x+9/4
(y+7/2) (y+7/2)= y^2 +7/2y +7/2y +49/4= y^2+7y+49/4

So, x^2+y^2-3x+7y-29/2= -sqrt(17/2)

I am doing independent study on a computer program. I have the values for the x and y coordinates right, but my value for the radius (sqrt(17/2)) is not right. I do not understand this because when I square this radius and move it over to the other side I get -6 just like the original equation of x^2 + y^2 - 3x + 7y - 6 = 0. What am I doing wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 21, 2010

LCKurtz

Try arranging your work like this:

x2 - 3x + y2 + 7y = 6