 # Find the required information and graph: displaystyle{2}{x}^{2}+{2}{y}^{2}+{2}{x}+{14}{y}+{17}={0} ankarskogC 2021-02-25 Answered
Find the required information and graph:
$2{x}^{2}+2{y}^{2}+2x+14y+17=0$
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Step 1
Consider the provided equation,
$2{x}^{2}+2{y}^{2}+2x+14y+17=0$
Classify the conic section and find the center.
We can write as,
$2{x}^{2}+2x+2{y}^{2}+14y+17=0$
$2\left({x}^{2}+x\right)+2\left({y}^{2}+7y\right)=-17$
$\left({x}^{2}+x\right)+\left({y}^{2}+7y\right)=-\frac{17}{2}$
$\left({x}^{2}+x+\frac{1}{4}\right)+\left({y}^{2}+7y+\frac{49}{4}\right)=-\frac{17}{2}+\frac{49}{4}+\frac{1}{4}$
Step 2
Simplifying further,
${\left(x+\frac{1}{2}\right)}^{2}+{\left(y+\frac{7}{2}\right)}^{2}=4$
${\left(x-\left(-\frac{1}{2}\right)\right)}^{2}+{\left(y-\left(-\frac{7}{2}\right)\right)}^{2}={2}^{2}$
${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}={r}^{2}$ is the circle equation with radius r, centered at (a, b).
Thus, this is a circle.
So, the center or the circle $\left(-\frac{1}{2},-\frac{7}{2}\right).$

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