How do you convert $(x-3)^2+(y+1)^2=10$ to polar form?

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Answer 1 · 2018-03-09T16:25:35

Polar form of equation is $r=6costheta-2sintheta$

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

$x=rcostheta$ , $y=rsintheta$ and hence $x^2+y^2=r^2$

Hence we can write $(x-3)^2+(y+1)^2=10$ as

$(rcostheta-3)^2+(rsintheta+1)^2=10$

or $r^2cos^2theta-6rcostheta+9+r^2sin^2theta+2rsintheta+1=10$

or $r^2-2r(3costheta-sintheta)=0$

or $r=6costheta-2sintheta$

The graph using tool at http://www.wolframalpha.com/ is shown below.

enter image source here

How do you convert (x-3)^2+(y+1)^2=10(x-3)^2+(y+1)^2=10 to polar form?