How do you convert #r=4 sin theta# to rectangular form?

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Answer 1 · 2016-05-04T08:45:29

#x^2+y^2-4y=0#. In the standard form, this is #x^2+(y-2)^2=2^2#

#r = 4 sin theta# represents the circle of diameter 4 and center at #

(2, pi/2)#.

For conversion to cartesian form, use #sin theta = y/r# and

#r^2=x^2+y^2#.

Substitutions give #r=4(y/r)#

At the pole #r=theta=0#, and so, x = y = 0. Elsewhere, cross-

multiplying, #r^2=x^2+y^2=4y#.

In the standard form, this is #x^2+(y-2)^2=2^2#.

Having noted that there were 8K viewers for this answer, I add

now more details.

graph{(x^2+(y-2)^2-2^2)(x^2+(y-2)^2-0.027)=0}
The general equation to circles passing through r = 0, with radius

'a' and center at polar #( a, alpha)# is

#r = 2a cos (theta - alpha)#.

Here, a = 2 and #alpha = pi/2#, to give #r = 4 sin theta#.

The circle for a = 2 and #alpha = pi/4# is shown, in the graph

below.

graph{((x-1.415)^2+(y-1.414)^2-4)((x-1.414)^2+(y-1.414)^2-0.027)=0}