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

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Answer 1 · 2016-07-03T10:05:33

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

Simplifying the given eqn., it becomes $: x^2+y^2+6x-4y+4=0.$

To convert this cartsn. eqn. into polar, we use the conversion formula $;x=rcostheta, y=rsintheta, and, x^2+y^2=r^2$

Sub.ing, we get the eqn. in polar form as $: r^2+2r(3costheta-2sintheta)+4=0$ .

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