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

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Answer 1 · 2016-09-18T05:03:37

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

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

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

$r = - 2 sin θ = - \frac{y}{r}$ $r=-2 sin theta =-y/r$ . and so,

$r^{2} = x^{2} + y^{2} = - 2 y$ $r^2=x^2+y^2=-2y$ .

In the standard form,

$x^{2} + {(y + 1)}^{2} = 1$ $x^2+(y+1)^2=1$ .

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

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

alpha)#. |d| is the diameter.

The rectangular form is

${(x + \frac{d}{2} cos α)}^{2} + {(y + \frac{d}{2} sin α)}^{2} = \frac{d^{2}}{4}$ $(x+d/2cos alpha)^2+(y+d/2sin alpha)^2=d^2/4$

Here, $d = - 2 and α = - \frac{π}{2}$ $d = -2 and alpha = -pi/2$ .. .

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