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

1 Answer
Shiva Prakash M V
Feb 12, 2018

#(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

Related questions
See all questions in Converting Between Systems
Impact of this question
946 views around the world
You can reuse this answer
Creative Commons License