# Thread: Equation of a Circle HELP!!!

1. ## Equation of a Circle HELP!!!

I have not had Math in 4 years! Can somone please show me how to do this problem:

The equation of a circle is 4x^2+4y^2+32x-32y+92=0
a. Finde the center (h,k) of the circle (50%)
b. Find the radius r of the circle (50%)

a: (h,k) =

b: r=

Thank You So Much!

2. the goal is to get the equation in the following form ...

$\displaystyle (x-h)^2 + (y-k)^2 = r^2$

where (h,k) are the center coordinates and r is the radius.

$\displaystyle 4x^2 + 4y^2 + 32x - 32y + 92 = 0$

start by dividing every term by 4 ...

$\displaystyle x^2 + y^2 + 8x - 8y + 23 = 0$

group the x-vales and y-values together, move the constant ...

$\displaystyle x^2 + 8x + y^2 - 8x = -23$

you now want to complete the square for both x and y ... half the coefficient of each linear term, square the result, and add to both sides of the equation ...

$\displaystyle x^2 + 8x + 16 + y^2 - 8x + 16 = -23 + 16 + 16$

factor the quadratic trinomials of x and y ...

$\displaystyle (x + 4)^2 + (y - 4)^2 = 3^2$

now, do you "see" h, k, and r?

3. Would the correct answer be:

(h,k) = (-4,4)
r = 3

4. yes