How do you graph 4x^2 + 4y^2 +32x +16y +71 = 04x2+4y2+32x+16y+71=0?

1 Answer
Cesareo R.
May 24, 2016

The graphic is the circumference (x+4)^2+(y+2)^2=(3/2)^2(x+4)2+(y+2)2=(32)2

Explanation:

The circumference equation is
(x-x_0)^2+(y-y_0)^2=r^2(x−x0)2+(y−y0)2=r2 so calculating
4 x^2 + 4 y^2 + 32 x + 16 y + 71 - 4((x-x_0)^2+(y-y_0)^2-r^2)=04x2+4y2+32x+16y+71−4((x−x0)2+(y−y0)2−r2)=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(32+8x0)x+(16+8y0)y+71+4r2−4x20−4y20=0
solving the equations for x_0,y_0,rx0,y0,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

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