How do you convert $r=-4cos(theta)$ to rectangular form?

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Answer 1 · 2016-02-15T07:32:12

$x^2+y^2+4x=0$

from the given, $r=-4 cos theta$

multiplying both sides of the equation by $r$

$r*r=r*(-4 cos theta)$
$r^2=-4*r cos theta$

Using $r=sqrt(x^2+y^2)$ and $x=r cos theta$

it follows

$r^2=-4*r cos theta$

$(sqrt(x^2+y^2))^2=-4*x$

$x^2+y^2=-4*x$

$x^2+y^2+4x=0$

have a nice day, ... from the Philippines !

How do you convert r=-4cos(theta)r=-4cos(theta) to rectangular form?