How do you convert #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:

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#