Question

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

Answer 1

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

Explanation:

`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}