Chapter 9.2, Problem 84AYU

To determine

To show: The graph of the polar equation $r = - 2\text{a}\text{sin}\theta , a>0$, is a circle of radius $a$ with center at $\left(0, - \text{a}\right)$ in rectangular coordinates.

Explanation of Solution

Given:

It is asked to show that the graph of the polar equation $r = - 2\text{a}\text{sin}\theta$, $a>0$, is a circle of radius $a$ with center at $\left(0, - \text{a}\right)$ in rectangular coordinates.

It is a known fact that each point in the polar coordinate system is given by $\left(r, \theta \right)$ where $r$ is the distance from the pole (origin) to the point, and $\theta$ is the counter clockwise angle that is made with the point pole, and the positive $x$-axis.

To convert back and forth between polar and rectangular coordinates, we have the following formulas:

$\begin{array}{l}\text{cos}\theta = \frac{x}{r}\\ x = r \text{cos}\theta \\ \text{sin}\theta = \frac{y}{r}\\ y = r \text{sin}\theta \\ r^2 = {\left(r \text{cos}\theta \right)}^2 + {\left(r \text{sin}\theta \right)}^2 = x^2 + y^2\\ \frac{\text{sin}\theta }{\text{cos}\theta } = \frac{\frac{y}{r}}{\frac{x}{r}}\\ \tan \theta = \frac{y}{x}\end{array}$

And $\left(x, y\right) = \left(r \text{cos}\theta , r \text{sin}\theta \right)$

Consider $r = - 2\text{a}\text{sin}\theta$,

Let $\text{sin}\theta = \frac{y}{r}$, the equation $r = - 2\text{a}\text{sin}\theta$ becomes

$\begin{array}{l}r = - 2a\left(\frac{y}{r}\right)\\ r^2 = - 2ay\\ x^2 + y^2 = - 2ay\\ x^2 + y^2 + 2ay + a^2 = a^2\end{array}$

$. x^2 + {\left(y + a\right)}^2 = a^2$

The equation becomes $x^2 + {\left(y + a\right)}^2 = a^2.$

In rectangular coordinates, $x^2 + {\left(y + a\right)}^2 = a^2$ is a circle of radius $a$ with center at $\left(0, - \text{a}\right)$ in rectangular coordinates.

In polar coordinates, $r = - 2\text{a}\text{sin}\theta$ is a circle of radius $a$ with center at $\left(0, - \text{a}\right)$ in rectangular coordinates.

Hence proved.