Question 661589: Please help me figure out this problem: 2x^2+2y^2+16x+8y−32=0 is the equation of a circle with center (h,k) and radius r for: h, k, and r
I got 2(x^2-8x+16) + 2 (y^2 + 4y + 4) = 32 +16 +4
divide by 2: (x-4)^2 + (y+2)^2 = 26
(4, -2) and radius of sqrt 26 or 5.09901951
my homework page has confirmed that only -2 was correct. Where did I make a mistake?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! standard form of equation for a circle:
(x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius
..
2x^2+2y^2+16x+8y−32=0
complete the square:
2x^2+16x+2y^2+8y−32=0
2(x^2+8x+16)+2(y^2+4y+4)=32+32+8
2(x+4)^2+2(y+2)^2=72
divide by 2
(x+4)^2+(y+2)^2=36 (equation of given circle)
h=-4
k=-2
r=6
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