Find the center (h, k) and the radius r of the circle 4 x^2 + 7 x + 4 y^2 - 6 y - 8 = 0.

To accomplish this we must rewrite 4 x^2 + 7 x + 4 y^2 - 6 y - 8 = 0 in standard form (x-h) ^2 + (y-k) ^2 = r^2.

How do we make 4 x^2 + 7 x + 4 into a perfect square? Try "completing the square."

4 x^2 + 7 x y^2 - 6 y - 8 = 0

Factor out the 4:

4 (x^2 + (7/4) x

Take half of the coeff. (7/4) of x and square it: (49/81)

Then we have

4 (x^2 + (7/4) + (49/81 - (49/81) + 4y^2 - 6y - 8 = 0

Then:

4 (x + 7/8) ^2 + 4y^2 - 6y - (49/4) - 8 = 0

4 (x+7/8) ^2 + 4 (y-3/4) ^2 - (9/16) - 8 = 0

4 (x+7/8) ^2 + 4 (y-3/4) ^2 - (9/16) - 8 = 0

4 (x+7/8) ^2 + 4 (y-3/4) ^2 - 8 9/16 = 0

4 (x+7/8) ^2 + 4 (y-3/4) ^2 = 8 9/16 = 137/16

Dividing all terms by 4, we get

(x+7/8) ^2 + (y-3/4) ^2 = 137/16

The radius is sqrt (137/16), or (1/4) sqrt (137). Center is (-7/8, 3/4).