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

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

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Answer 1 · 2016-07-21T08:05:11

#r^2-8rcostheta-10rsintheta+37=0#

Explanation:

Relation between polar coordinates #(r,theta)# and rectangular coordinates is given by #x=rcostheta#, #y=rsintheta#, #r^2=x^2+y^2# and #tantheta=y/x#.

Hence #4=(x-4)^2+(y-5)^2# can be written as

#x^2-8x+16+y^2-10y+25=4# or

#x^2+y^2-8x-10y+37=0# or

#r^2-8rcostheta-10rsintheta+37=0#

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

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