Expert: Ahmed Salami - 5/27/2010

Question
how do i solve for the radius and the center of a circle when i have
X^2+Y^2-8X+4Y-8=0?

Answer
Hi Diana,
The equation of a circle is of the form
(x - h)² + (y - k)² = r²
where the center is (h, k) and the radius is r
You need to be familiar with the method of completing squares. Lets say we want to complete the square of x² + Kx, we need to add (K/2)² and then factorize as
x² + Kx + (K/2)² = (x + K/2)²
Now, the equation of a circle involves both x and y terms and so we complete squares both in x and y remembering to preserve the equality sign by always adding equal amounts to both sides as you will see.
x² + y² - 8x + 4y - 8 = 0
First, we re-arrange
x² - 8x + y² + 4y = 8
Adding values to complete squares in x and y (remember to add to both sides)
x² - 8x + (-8/2)² + y² + 4y + (4/2)² = 8 + (-8/2)² + (4/2)²
x² - 8x + (-4)² + y² + 4y + (2)² = 8 + (-4)² + (2)²
(x - 4)² + (y + 2)² = 8 + 16 + 4
(x - 4)² + (y + 2)² = 28
(x - 4)² + (y + 2)² = (2√7)²
Comparing with the general equation, we can see that the center is at (4, -2) while the radius is 2√7.

Regards