Question

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

Answer 1

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).