How do you convert $9=(x-7)^2+(y+7)^2$ into polar form?

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Answer 1 · 2017-07-15T17:18:42

$r^2-14r(sintheta-costheta)+89=0$

or $r^2-14sqrt2rsin(theta-pi/4)+89=0$

The relation between polar coordinates $(r,theta)$ and rectangular coordinates $(x,y)$ is given by

$x=rcostheta$ , $y=rsintheta$ i.e. $x^2+y^2=r^2$

Hence we can write $9=(x-7)^2+(y+7)^2$ as

$(rcostheta-7)^2+(rsintheta+7)^2=9$

or $r^2cos^2theta-14rcostheta+49+r^2sin^2theta+14rsintheta+49=9$

or $r^2-14r(sintheta-costheta)+89=0$

or $r^2-14sqrt2rsin(theta-pi/4)+89=0$

How do you convert 9=(x-7)^2+(y+7)^29=(x-7)^2+(y+7)^2 into polar form?