How do you convert $9=(x-6)^2+(y+2)^2$ into polar form?

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Answer 1 · 2016-08-03T13:02:42

$r^2-4r(3costheta-sintheta)+31=0$ .

We, first simplify the given eqn., and, get

$x^2+y^2-12x+4y+31=0..................(1)$ .

For he conversion from cartesian to polar, we need the following

formulas $: x=rcostheta, y=rsintheta$ .

Sub.ing, in $(1)$ , we have,

$r^2-4r(3costheta-sintheta)+31=0$ .

How do you convert 9=(x-6)^2+(y+2)^29=(x-6)^2+(y+2)^2 into polar form?