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

1 Answer
May 24, 2016

Answer:

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#