How do you convert #x^2+(y-4)^2=16# to polar form?

2 Answers
Shwetank Mauria
May 3, 2016

#r=8sintheta#

Explanation:

If #(r,theta)# is in polar form and #(x,y)# in Cartesian form the relation between them is as follows:

#x=rcostheta#, #y=rsintheta#, #r^2=x^2+y^2# and #tantheta=y/x#

Hence, #x^2+(y-4)^2=16# can be written as

#x^2+y^2-8y+16=16# or

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

#r^2-8rsintheta=0# or

#r(r-8sintheta)=0# dividing by #r#

#r-8sintheta=0# or

#r=8sintheta#

Jim G.
May 3, 2016

#r=8sintheta#

Explanation:

To convert from Cartesian to Polar coordinates use the following formulae that link them.

#• x=rcostheta" and " y=rsintheta#

#x^2+y^2-8y+16=16 " (expanding bracket) "#

#rArrr^2cos^2theta+r^2sin^2theta-8rsintheta+16-16=0#

then #r^2(cos^2theta+sin^2theta)-8rsintheta=0#

using the identity: #cos^2theta+sin^2theta=1 #

#rArr r^2=8rsintheta" and dividing both sides by r "#

#rArr r=8sintheta#

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