How do you convert #4=(x-3)^2+(y-5)^2# into polar form?
1 Answer
May 2, 2016
Explanation:
To convert into polar form, substitute:
#{ (x = r cos theta), (y = r sin theta) :}#
#4 = (r cos theta - 3)^2+(r sin theta - 5)^2#
#=r^2 cos^2 theta - 6r cos theta + 9 + r^2 sin^2 theta - 10r sin theta + 25#
#=r^2 (cos^2 theta + sin^2 theta) - 2r(3cos theta + 5 sin theta)+34#
#=r^2 - 2r(3 cos theta + 5 sin theta) + 34#
Subtract
#r^2 - 2r(3 cos theta + 5 sin theta) + 30 = 0#
