Respuesta :

Answer:

[tex]r^2-12r\sin\theta=1[/tex]

Explanation:

We need to write the polar form of the given equation :

[tex]x^2+(y-6)^2=36[/tex]

Put [tex]x=r\cos\theta[/tex] and [tex]y=r\sin\theta[/tex]

[tex]x^2+(y-6)^2=36\\\\(r\cos\theta)^2+(r\sin\theta-6)^2=36\\\\r^2\cos^2\theta+r^2\sin^2\theta+36-12r\sin\theta=36\ \ [\because (a-b)^2=a^2+b^2-2ab]\\\\r^2(\cos^2\theta+\sin^2\theta)-12r\sin\theta=0\\\\\text{We know that}, \cos^2\theta+\sin^2\theta=1\\\\r^2-12r\sin\theta=1[/tex]

Hence, the above is the polar form of the given equation i.e. [tex]r^2-12r\sin\theta=1[/tex]

Answer:

I am assuming it is D. r = 12sin(θ)

Explanation:

Update just finished my quiz and got it correct, it is D!

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