# 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

Oct 28, 2016

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

radius $r = \sqrt{\frac{165}{16}} = \frac{\sqrt{165}}{4}$

#### Explanation:

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

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.

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

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

(x^2+5x+(5/2)^2)+(y^2+y/2+(1/4)^2)-4-(5/2)^2-(1/4)^"=0

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

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

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

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

radius $r = \sqrt{\frac{165}{16}} = \frac{\sqrt{165}}{4}$