How do you convert $r = 4 cos (θ)$ $r = 4cos(theta)$ into rectangular form?

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How do you convert $r = 4 cos (θ)$ $r = 4cos(theta)$ into rectangular form?

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Answer 1 · 2016-03-08T18:42:20

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

Explanation:

$r = 4 cos θ$ $r=4costheta$
$r^{2} = 4 r cos θ$ $r^2=4rcos theta$ -> multiply both sides by r
$x^{2} + y^{2} = 4 x$ $x^2+y^2 = 4x$ -> substitute in $x^{2} + y^{2}$ $x^2+y^2$ for $r^{2}$ $r^2$ and $x$ $x$ for $r cos θ$ $rcos theta$
$x^{2} - 4 x + y^{2} = 0$ $x^2-4x+y^2=0$ -> combine like terms
${(x - 2)}^{2} + y^{2} = 4$ $(x-2)^2+y^2=4$ ->complete the square

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How do you convert $r = 4 cos (θ)$ $r = 4cos(theta)$ into rectangular form?

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How do you convert $r = 4 cos (θ)$ $r = 4cos(theta)$ into rectangular form?