How do I factor this with the 2x^2 ?

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How do I factor this with the 2x^2 ?

Find the center ( h,k ) and the radius of the circle 2x^2 + 10x +2y^2 + y - 8 = 0

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Answer 1 · 2016-10-28T15:34:52

centre is $(- \frac{5}{2}, - \frac{1}{4})$ $(-5/2, -1/4)$

radius $r = \sqrt{\frac{165}{16}} = \frac{\sqrt{165}}{4}$ $r=sqrt(165/16)=sqrt165/4$

Explanation:

$2 x^{2} + 10 x + 2 y^{2} + y - 8 = 0$ $2x^2+10x+2y^2+y-8=0$

to find centre of the circle we need to complete the squares on the x & y parts of the equation.

When completing the square get the coefficient of the square terms =1

Divide everything by 2 to begin with.

$x^{2} + 5 x + y^{2} + \frac{y}{2} - 4 = 0$ $x^2+5x+y^2+y/2-4=0$

$(x^{2} + 5 x) + (y^{2} + \frac{y}{2}) - 4 = 0$ $(x^2+5x)+(y^2+y/2)-4=0$

$(x^{2} + 5 x + {(\frac{5}{2})}^{2}) + (y^{2} + \frac{y}{2} + {(\frac{1}{4})}^{2}) - 4 - {(\frac{5}{2})}^{2} - {(\frac{1}{4})} = 0$ $(x^2+5x+(5/2)^2)+(y^2+y/2+(1/4)^2)-4-(5/2)^2-(1/4)^"=0$

${(x + \frac{5}{2})}^{2} + {(y + \frac{1}{4})}^{2} - 4 - \frac{25}{4} - \frac{1}{16} = 0$ $(x+5/2)^2+(y+1/4)^2-4-25/4-1/16=0$

${(x + \frac{5}{2})}^{2} + {(y + \frac{1}{4})}^{2} = \frac{165}{16}$ $(x+5/2)^2+(y+1/4)^2=165/16$

cmp ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$

centre is $(- \frac{5}{2}, - \frac{1}{4})$ $(-5/2, -1/4)$

radius $r = \sqrt{\frac{165}{16}} = \frac{\sqrt{165}}{4}$ $r=sqrt(165/16)=sqrt165/4$

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How do I factor this with the 2x^2 ?

Find the center ( h,k ) and the radius of the circle 2x^2 + 10x +2y^2 + y - 8 = 0

1 Answer

Explanation:

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