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

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Answer 1 · 2016-05-22T12:54:54

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

$r^2+2r(3costheta+8sintheta)+64=0$

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

$x=rcostheta$ and $y=rsintheta$ and hence

$x^2+y^2=r^2cos^2theta+r^2sin^2theta=r^2$

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

$(rcostheta+3)^2+(rsintheta+8)^2=9$

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

or $r^2+r(6costheta+16sintheta)+64=0$

or $r^2+2r(3costheta+8sintheta)+64=0$

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