How do you convert #r= -2sin(theta)# into rectangular form?

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Answer 1 · 2016-09-18T05:03:37

#x^2+(y+1)^2=1#.
This represents the unit circle through the orifgin, with center at (0, -1).

Use the conversion equation #r(cos theta, sin theta)=(x, y)#.

As #y =r sin theta and r#2=x^2+y^2#,

#r=-2 sin theta =-y/r#. and so,

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

In the standard form,

#x^2+(y+1)^2=1#.

This represents the unit circle through the origin with center at (0, -1).

The general polar equation for the family of circles through the

pole r = 0 is

#r = dcos(theta+alpha)# that has center at #(d/2 cos alpha, d/2sin

alpha)#. |d| is the diameter.

The rectangular form is

#(x+d/2cos alpha)^2+(y+d/2sin alpha)^2=d^2/4#

Here, #d = -2 and alpha = -pi/2#.. .