How do you convert $4 = {(x + 8)}^{2} + {(y - 5)}^{2}$ $4=(x+8)^2+(y-5)^2$ into polar form?

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Answer 1 · 2016-03-20T12:42:51

Set:

$x=rcosθ$

$y=rsinθ$

Answer is:

$r^2+r*(16cosθ-10sinθ)+85=0$

According to the geometry of this picture:

![http://mathinsight.org/]( useruploads.socratic.org )

Set:

$x=rcosθ$

$y=rsinθ$

Substitute into the equation:

$4=(x+8)^2+(y-5)^2$

$4=(rcosθ+8)^2+(rsinθ-5)^2$

$4=color(red)(r^2cos^2θ)+16*rcosθ+color(green)(64)+color(red)(r^2sin^2θ)-10*rsinθ+color(green)(25)$

$color(purple)(4)=r^2*color(blue)((cos^2θ+sin^2θ))+16*rcosθ-10*rsinθ+color(purple)(89)$

$0=r^2*1+color(red)(16*rcosθ-10*rsinθ)+85$

$r^2+r*(16cosθ-10sinθ)+85=0$

How do you convert 4=(x+8)^2+(y-5)^24=(x+8)2+(y−5)24=(x+8)^2+(y-5)^2 into polar form?