How do you write the equation using polar coordinates given ${(x - 2)}^{2} + y^{2} = 4$ $(x-2)^2+y^2=4$ ?

Question

1401 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-01T11:05:29

This is the equation of a circle with center in (2,0) and radius 2 and becomes:

$ρ^{2} - 4 ρ cos θ = 0$ $rho^2 -4rho cos theta =0$

Substitute the expression of $x$ $x$ and $y$ $y$ in terms of polar coordinates:

$x = ρ cos θ$ $x=rho cos theta$
$y = ρ sin θ$ $y=rho sin theta$

and you have:

${(ρ cos θ - 2)}^{2} + ρ^{2} {sin}^{2} θ = 4$ $(rho cos theta - 2)^2+rho^2 sin^2 theta = 4$

$rho^2cos^2theta -4rho cos theta +cancel 4 + rho^2 sin^2 theta = cancel 4$

$rho^2(cos^2theta +sin^2 theta) -4rho cos theta =0$

$rho^2 -4rho cos theta =0$

How do you write the equation using polar coordinates given (x-2)^2+y^2=4(x−2)2+y2=4(x-2)^2+y^2=4?