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).
