How do you convert $4 = {(x + 8)}^{2} + {(y - 6)}^{2}$ $4=(x+8)^2+(y-6)^2$ into polar form?

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Answer 1 · 2016-05-23T09:04:06

Polar form is $r^{2} + 4 r (4 cos θ - 3 sin θ) + 96 = 0$ $r^2+4r(4costheta-3sintheta)+96=0$

A Cartesian point $(x, y)$ $(x,y)$ in polar form is $(r, θ)$ $(r,theta)$ , where

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

$x^{2} + y^{2} = r^{2} {cos}^{2} θ + r^{2} {sin}^{2} θ = r^{2}$ $x^2+y^2=r^2cos^2theta+r^2sin^2theta=r^2$

Hence $4 = {(x + 8)}^{2} + {(y - 6)}^{2}$ $4=(x+8)^2+(y-6)^2$ can be written as

${(r cos θ + 8)}^{2} + {(r sin θ - 6)}^{2} = 4$ $(rcostheta+8)^2+(rsintheta-6)^2=4$

or $r^{2} {cos}^{2} θ + 16 r cos θ + 64 + r^{2} {sin}^{2} θ - 12 r sin θ + 36 = 4$ $r^2cos^2theta+16rcostheta+64+r^2sin^2theta-12rsintheta+36=4$

or $r^{2} + r (16 cos θ - 12 sin θ) + 64 + 36 - 4 = 0$ $r^2+r(16costheta-12sintheta)+64+36-4=0$

or $r^{2} + 4 r (4 cos θ - 3 sin θ) + 96 = 0$ $r^2+4r(4costheta-3sintheta)+96=0$

How do you convert 4=(x+8)^2+(y-6)^24=(x+8)2+(y−6)24=(x+8)^2+(y-6)^2 into polar form?