anonymous
  • anonymous
Find the center (h, k) and the radius r of the circle 2 x^2 - 6 x +2 y^2 + 9 y - 3 = 0 .
Mathematics
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anonymous
  • anonymous
Find the center (h, k) and the radius r of the circle 2 x^2 - 6 x +2 y^2 + 9 y - 3 = 0 .
Mathematics
katieb
  • katieb
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zzr0ck3r
  • zzr0ck3r
You need to complete the square twice. I will do one of them, \[2 x^2 - 6 x +2 y^2 + 9 y - 3 = 0\\2x^2-6x+2y^2+9y=3\\2(x^2+3x)+2y^2+9y=3\\2(x+\frac{3}{2})^2+2y^2+9y=3+2(\frac{3}{2})^2\]
anonymous
  • anonymous
why is it 3/2?
zzr0ck3r
  • zzr0ck3r
Equation of a circle is \[(x-h)^2+(y-k)^2=r^2\] Where \((h,k)\) is the center and \(r\) is the radius. Yuu have to complete the square twice. \[2 x^2 - 6 x +2 y^2 + 9 y - 3 = 0 \\2x^2-6x+2y^2+9y=3 \\2(x^2-3x)+2(y^2+\frac{9}{2}y)=3 \\2(x-\frac{3}{2})^2+2(y+\frac{9}{4})^2=3+2(\frac{3}{2})^2+2(\frac{9}{4})^2=\frac{141}{8}\\\\(x-\frac{3}{2})^2+(y+\frac{9}{4})^2=\frac{141}{16}\]

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anonymous
  • anonymous
ohhh okay that makes a lot of sense..
anonymous
  • anonymous
wait, ive tried it again but for radius i got 117/8. I tried to enter both but none of the radius were right @zzr0ck3r
zzr0ck3r
  • zzr0ck3r
The radius would be, according to the formula, \(r=\sqrt{\frac{141}{16}}=\frac{\sqrt{141}}{4}\).

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