Question #efd99

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Answer 1 · 2016-10-09T14:43:08

We know the relations

$x = r cos θ and y = r sin θ$ $x=rcostheta and y =rsintheta$

Again $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

where r and $θ$ $theta$ are the polar coordinate of a point having rectangular coordinate $(x, y)$ $(x,y)$

The given equation in rectanglar form is

${(x + 2)}^{2} + y^{2} = 4$ $(x+2)^2+y^2=4$

$\Rightarrow x^{2} + 4 x + 4 + y^{2} = 4$ $=>x^2+4x+4+y^2=4$

$\Rightarrow x^{2} + y^{2} + 4 x = 0$ $=>x^2+y^2+4x=0$

$\Rightarrow r^{2} + 4 r cos θ = 0$ $=>r^2+4rcostheta=0$

$\Rightarrow r + 4 cos θ = 0$ $=>r+4costheta=0$

This is the polar form of the given equation.