How do you convert $9 = {(x + 3)}^{2} + {(y + 8)}^{2}$ $9=(x+3)^2+(y+8)^2$ into polar form?

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Answer 1 · 2016-05-22T12:54:54

$9 = {(x + 3)}^{2} + {(y + 8)}^{2}$ $9=(x+3)^2+(y+8)^2$ in polar form can be written as

$r^{2} + 2 r (3 cos θ + 8 sin θ) + 64 = 0$ $r^2+2r(3costheta+8sintheta)+64=0$

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 $9 = {(x + 3)}^{2} + {(y + 8)}^{2}$ $9=(x+3)^2+(y+8)^2$ can be written as

${(r cos θ + 3)}^{2} + {(r sin θ + 8)}^{2} = 9$ $(rcostheta+3)^2+(rsintheta+8)^2=9$

or $r^{2} {cos}^{2} θ + 6 r cos θ + 9 + r^{2} {sin}^{2} θ + 16 r sin θ + 64 = 9$ $r^2cos^2theta+6rcostheta+9+r^2sin^2theta+16rsintheta+64=9$

or $r^{2} + r (6 cos θ + 16 sin θ) + 64 = 0$ $r^2+r(6costheta+16sintheta)+64=0$

or $r^{2} + 2 r (3 cos θ + 8 sin θ) + 64 = 0$ $r^2+2r(3costheta+8sintheta)+64=0$

How do you convert 9=(x+3)^2+(y+8)^29=(x+3)2+(y+8)29=(x+3)^2+(y+8)^2 into polar form?