How do you convert $r = 12 cos [θ]$ $r=12 cos [theta]$ to rectangular form?

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Answer 1 · 2016-11-25T17:39:38

The equation becomes a circle with radius 6 and centered at $(6, 0)$ $(6, 0)$ : ${(x - 6)}^{2} + {(y - 0)}^{2} = 6^{2}$ $(x - 6)^2 + (y - 0)^2 = 6^2$

Multiply both sides by r:

$r^{2} = 12 r cos (θ)$ $r^2 = 12rcos(theta)$

Substitute $x^{2} + y^{2}$ $x^2 + y^2$ for $r^{2}$ $r^2$ and x for $r cos (θ)$ $rcos(theta)$

$x^{2} + y^{2} = 12 x$ $x^2 + y^2 = 12x$

Subtract 12x from both sides:

$x^{2} - 12 x + y^{2} = 0$ $x^2 - 12x + y^2 = 0$

Add h^2 to both sides:

$x^{2} - 12 x + h^{2} + y^{2} = h^{2}$ $x^2 - 12x + h^2 + y^2 = h^2$

Complete the square for #(x - h)^2 = x^2 - 2hx + h^2:

-2hx = -12x

$h = 6$ $h = 6$

Insert the square term with h = 6 on the left and substitute 6 for h on the right:

${(x - 6)}^{2} + y^{2} = 6^{2}$ $(x - 6)^2 + y^2 = 6^2$

Write the y term is standard form:

${(x - 6)}^{2} + {(y - 0)}^{2} = 6^{2}$ $(x - 6)^2 + (y - 0)^2 = 6^2$

Answer 2 · 2016-11-25T17:40:47

$x^{2} + y^{2} - 12 x = 0$ $x^2+y^2-12x= 0$

Multiply the given expression on both sides by r, it becomes $r^{2} = 12 r cos θ$ $r^2= 12 r cos theta$

Substitute $r cos θ = x$ $rcos theta =x$ and $r^{2} = x^{2} + y^{2}$ $r^2= x^2+y^2$

The equation now becomes, $x^{2} + y^{2} = 12 x$ $x^2+y^2 = 12x$
$x^{2} + y^{2} - 12 x = 0$ $x^2 +y^2 -12x=0$

How do you convert r=12 cos [theta]r=12cos[θ]r=12 cos [theta] to rectangular form?