How do you convert $r = - 9 cos θ$ $r = -9 cos theta$ into rectangular form?

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Answer 1 · 2015-11-26T12:11:53

$x^{2} + 9 x + y^{2} = 0$ $x^2+9x+y^2=0$

Since $r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$ , if you multiply both sides by $r$ $r$ you have

$r^{2} = - 9 r cos (θ)$ $r^2 = -9rcos(theta)$

Now, $r^{2} = x^{2} + y^{2}$ $r^2 = x^2+y^2$ , and $r cos (θ) = x$ $rcos(theta)=x$ .

So, the equation becomes

$x^{2} + y^{2} = - 9 x$ $x^2+y^2 = -9x$ , which is a circle.

How do you convert r = -9 cos thetar=−9cosθr = -9 cos theta into rectangular form?