Identify and graph the equations:

(a) $r\sin \theta =2$

(b) $r=4\sin \theta$

Figure 41 $r\sin \theta =2$ or $y=2$.

Solution (a) Here, both $r$ and $\theta$ are allowed to vary, so the graph of the equation is not as obvious. If we use the fact that $y=r\sin \theta ,$ the equation $r\sin \theta =2$ becomes $y=2.$ So, the graph of $r\sin \theta =2$ is the horizontal line $y=2$ that lies $2$ units above the pole, as shown in Figure 41.

(b) To convert the equation $r=4\sin \theta$ to rectangular coordinates, we multiply the equation by $r$ to obtain $$\begin{equation*} r^{2}=4r\sin \theta \end{equation*}$$

Now we use the formulas $r^{2}=x^{2}+y^{2}$ and $y=r\sin \theta$. Then $$\begin{array}{rcl@{\qquad}l} r^{2} &=&4r\sin \theta \\ x^{2}+y^{2} &=&4y & {\color{#0066A7}{{\hbox{\(r^{2} =x^{2} + y^{2},\;r\sin \theta =y\)}}}} \\ x^{2}+(y^{2}-4y) &=&0 \\ x^{2}+\left( y^{2}-4y+4\right) &=&4 & {\color{#0066A7}{{\hbox{Complete the square in \(y\).}}}} \\ x^{2}+(y-2)^{2} &=&4 & {\color{#0066A7}{{\hbox{Factor.}}}} \end{array}$$

Figure 42 $r=4\sin \theta$ or $x^{2}+(y-2)^{2}=4$.

This is the standard form of the equation of a circle with its center at the point $( 0,2)$ and radius $2$ in rectangular coordinates. See Figure 42. Notice that the circle passes through the pole.