How do you convert $9 = {(x - 3)}^{2} + {(2 y - 9)}^{2}$ $9=(x-3)^2+(2y-9)^2$ into polar form?

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Answer 1 · 2018-05-03T00:35:27

The conversion from Rectangular to Polar:
$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

Substitute for $x$ $x$ and $y$ $y$ :
$9 = {(r cos θ - 3)}^{2} + {(2 r sin θ - 9)}^{2}$ $9=(rcostheta-3)^2+(2rsintheta-9)^2$

$r^{2} {cos}^{2} θ - 6 r cos θ + 9 + 4 r^{2} {sin}^{2} θ - 36 {sin}^{2} θ + 81 = 9$ $r^2cos^2theta-6rcostheta+9+4r^2sin^2theta-36sin^2theta+81=9$

$r^{2} {cos}^{2} θ - 6 r cos θ + 4 r^{2} {sin}^{2} θ - 36 r sin θ = - 81$ $r^2cos^2theta-6rcostheta+4r^2sin^2theta-36rsintheta=-81$

$r (r {cos}^{2} θ - 6 cos θ + 4 r {sin}^{2} θ - 36 sin θ) = - 81$ $r(rcos^2theta-6costheta+4rsin^2theta-36sintheta)=-81$

$r = - 81$ $r=-81$

Or the more meaningful solution:

$r {cos}^{2} θ - 6 cos θ + 4 r {sin}^{2} θ - 36 sin θ = - 81$ $rcos^2theta-6costheta+4rsin^2theta-36sintheta=-81$

$r {cos}^{2} θ + 4 r {sin}^{2} θ = - 81 + 6 cos θ + 36 sin θ$ $rcos^2theta+4rsin^2theta=-81+6costheta+36sintheta$

$r ({cos}^{2} θ + 4 {sin}^{2} θ) = - 81 + 6 cos θ + 36 sin θ$ $r(cos^2theta+4sin^2theta)=-81+6costheta+36sintheta$

$r = \frac{- 81 + 6 cos θ + 36 sin θ}{{cos}^{2} θ + 4 {sin}^{2} θ}$ $r=(-81+6costheta+36sintheta)/(cos^2theta+4sin^2theta)$

How do you convert 9=(x-3)^2+(2y-9)^29=(x−3)2+(2y−9)29=(x-3)^2+(2y-9)^2 into polar form?