How do you convert $4 = {(x - 2)}^{2} + {(y - 6)}^{2}$ $4=(x-2)^2+(y-6)^2$ into polar form?

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How do you convert $4 = {(x - 2)}^{2} + {(y - 6)}^{2}$ $4=(x-2)^2+(y-6)^2$ into polar form?

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Answer 1 · 2016-01-28T14:53:17

$r = 2 cos θ + 6 sin θ \pm 2 \sqrt{3 sin 2 θ - 8 {cos}^{2} θ}$ $r=2costheta+6sintheta+-2sqrt(3sin2theta-8cos^2theta)$

Explanation:

Use the rule for switching from a Cartesian system to a polar system:

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

This gives:

$4 = {(r cos θ - 2)}^{2} + {(r sin θ - 6)}^{2}$ $4=(rcostheta-2)^2+(rsintheta-6)^2$

$4 = r^{2} {cos}^{2} θ - 4 r cos θ + 4 + r^{2} {sin}^{2} θ - 12 r sin θ + 36$ $4=r^2cos^2theta-4rcostheta+4+r^2sin^2theta-12rsintheta+36$

$0 = r^{2} {cos}^{2} θ + r^{2} {sin}^{2} θ - 4 r cos θ - 12 r sin θ + 36$ $0=r^2cos^2theta+r^2sin^2theta-4rcostheta-12rsintheta+36$

$0 = r^{2} ({cos}^{2} θ + {sin}^{2} θ) + r (- 4 cos θ - 12 sin θ) + 36$ $0=r^2(cos^2theta+sin^2theta)+r(-4costheta-12sintheta)+36$

Note that ${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta+sin^2theta=1$ through the Pythagorean identity.

$0 = r^{2} + r (- 4 cos θ - 12 sin θ) + 36$ $0=r^2+r(-4costheta-12sintheta)+36$

This can be solved through the quadratic equation.

$r = \frac{4 cos θ + 12 sin θ \pm \sqrt{16 {cos}^{2} θ + 96 cos θ sin θ + 144 {sin}^{2} θ - 144}}{2}$ $r=(4costheta+12sintheta+-sqrt(16cos^2theta+96costhetasintheta+144sin^2theta-144))/2$

$r = \frac{4 cos θ + 12 sin θ \pm 4 \sqrt{{cos}^{2} θ + 6 cos θ sin θ + 9 {sin}^{2} θ - 9}}{2}$ $r=(4costheta+12sintheta+-4sqrt(cos^2theta+6costhetasintheta+9sin^2theta-9))/2$

$r = 2 cos θ + 6 sin θ \pm 2 \sqrt{{cos}^{2} θ + 6 cos θ sin θ + 9 {sin}^{2} θ - 9}$ $r=2costheta+6sintheta+-2sqrt(cos^2theta+6costhetasintheta+9sin^2theta-9)$

$r = 2 cos θ + 6 sin θ \pm 2 \sqrt{{cos}^{2} θ + 6 cos θ sin θ + 9 ({sin}^{2} θ - 1)}$ $r=2costheta+6sintheta+-2sqrt(cos^2theta+6costhetasintheta+9(sin^2theta-1))$

$r = 2 cos θ + 6 sin θ \pm 2 \sqrt{{cos}^{2} θ + 6 cos θ sin θ - 9 {cos}^{2} θ}$ $r=2costheta+6sintheta+-2sqrt(cos^2theta+6costhetasintheta-9cos^2theta)$

$r = 2 cos θ + 6 sin θ \pm 2 \sqrt{6 cos θ sin θ - 8 {cos}^{2} θ}$ $r=2costheta+6sintheta+-2sqrt(6costhetasintheta-8cos^2theta)$

Note that $2 cos θ sin θ = sin 2 θ$ $2costhetasintheta=sin2theta$ , so $6 cos θ sin θ = 3 sin 2 θ$ $6costhetasintheta=3sin2theta$ .

$r = 2 cos θ + 6 sin θ \pm 2 \sqrt{3 sin 2 θ - 8 {cos}^{2} θ}$ $r=2costheta+6sintheta+-2sqrt(3sin2theta-8cos^2theta)$

If you have any questions as to any of the underlying algebra going on here, feel free to ask. I felt it would be a little much to try to explain each step.

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How do you convert $4 = {(x - 2)}^{2} + {(y - 6)}^{2}$ $4=(x-2)^2+(y-6)^2$ into polar form?

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How do you convert $4 = {(x - 2)}^{2} + {(y - 6)}^{2}$ $4=(x-2)^2+(y-6)^2$ into polar form?