How do you convert $4=(5x-3)^2+(y-5)^2$ into polar form?

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How do you convert $4=(5x-3)^2+(y-5)^2$ into polar form?

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Answer 1 · 2018-02-12T12:02:17

$(25cos^2theta+sin^2theta)r^2-(30costheta+10sintheta)r+30=0$ is the polar form

Explanation:

Given:
$4=(5x-3)^2+(y-5)^2$
$x=rcostheta$
$y=rsintheta$
Thus,
$4=(5rcostheta-3)^2+(rsintheta-5)^2$
$4=25r^2cos^2theta-30rcostheta+9+r^2sin^2theta-10rsintheta+25$
$(25cos^2theta+sin^2theta)r^2-(30costheta+10sintheta)r+9+25-4=0$
$(25cos^2theta+sin^2theta)r^2-(30costheta+10sintheta)r+30=0$ is the polar form

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How do you convert $4=(5x-3)^2+(y-5)^2$ into polar form?

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