Circles

We know (hopefully) that the equation x² + y² = a² represents a circle of radius a centered at the origin. We can rewrite this equation as

$$\left( \frac{x}{a} \right)^2 + \left( \frac{y}{a} \right)^2 = 1.$$ (18)

Noting that if we let

$$\left( \frac{x}{a} \right) = \cos t, \thinspace \mbox{and} \thinspace \left( \frac{y}{a} \right) = \sin t,$$ (19)

then the parameterization $\vec{r}(t) = a \cos t \hat{i} + a \sin t \hat{j}, 0 \le t \le 2\pi$ will trace out the circle. This particular parameterization starts at the point (a,0) and traces the curve counterclockwise as t increases. How could we start the circle at the point (0,1)? There are two obvious ways (and other not-as-obvious ways):

1.

change the range of t to go from $\pi/2 \le t 5\pi/2$, or

2.

change the parameter itself from t to $t + \pi/2$.

Of these choices, the second is probably better in the sense that the range of t starts at 0.

How can we trace the circle put clockwise instead of counter clockwise? One way is to let t become -t. Then the parameterization (using some properties of sine and cosine) becomes $\vec{r}(t) = a \cos t \hat{i} - a \sin t \hat{j}$.

How can we center the circle at a point $\vec{r}_0 = x_0 \hat{i} + y_0 \hat{j} + z_0 \hat{k}$? To do this, use the picture below to help with the vector addition. The result is $\vec{r}(t) = (x_0 + a\cos t)\hat{i} + (y_0 \vert a \sin t) \hat{j} + z_0 \hat{k}$.