How do you graph #r=4sintheta#?

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Answer 1 · 2016-12-17T01:55:41

Graph of this circle through the pole r = 0, with radius 2 and center at (2, 1.57), nearly, is inserted

The equation #r=2a cos (theta-alpha)# represents the family of

circles through the pole r = 0 and having center on #theta=alpha# at

#(a, alpha)#.

Here, #a = 2 and alpha =pi/2# and the center is at #(1, pi/2)#.

In cartesian form,

#r = sqrt(x^2+y^2)=4 sin theta = 4(y/r)=4(y/sqrt(x^2+y^2))#

Graph for this is inserted.

graph{(x^2+y^2)-4x=0 [-5.64, 5.64, -2.82, 2.82]}