Question

How do you graph `4x^2 + 4y^2 +32x +16y +71 = 0`?

Answer 1

The graphic is the circumference `(x+4)^2+(y+2)^2=(3/2)^2`

Explanation:

The circumference equation is
`(x-x_0)^2+(y-y_0)^2=r^2` so calculating
`4 x^2 + 4 y^2 + 32 x + 16 y + 71 - 4((x-x_0)^2+(y-y_0)^2-r^2)=0` we get
`(32 + 8 x_0)x+(16 + 8 y_0)y + 71 + 4 r^2 - 4 x_0^2 - 4 y_0^2=0`
solving the equations for `x_0,y_0,r`
`((32 + 8 x_0=0),(16 + 8 y_0=0),(71 + 4 r^2 - 4 x_0^2 - 4 y_0^2=0))`
we get
`x_0 = -4, y_0 = -2, r = 3/2`