A unit circle is a circle with a radius of \(1\). Because every circle has a radius, all other circles are just some magnification of the unit circle. Do note, however, that the area will not double if the radius doubles. To see why, consider the area formula:
$$A = \pi r^2$$
If we double the radius, we substitute \(r \rightarrow 2r\):
$$A = \pi (2r)^2 = 4\pi r^2 = 4A$$
In other words, the area will quadruple.
If we leave a mark on a circle of some known radius and rotate it like a wheel a known number of degrees, we can find the position of the mark using the same concepts from using a unit circle.
Example 1: We have a wheel of radius 2 meters and mark the rightmost point on the circle with a red dot. We spin the wheel \(240\) degrees clockwise. If the center of the wheel is at \((0, 0)\), at what point on the Cartesian Plane is the red dot now at?
Solution: When we travel along the unit circle, we always start at the rightmost part on the circle. This example is no exception, but note the radius of the circle is \(2\), so we start at point \((2, 0)\). All points are twice as far away from the origin as they would be on a unit circle, so a point on the unit circle \((\cos \theta, \sin \theta)\) corresponds to \((2\cos \theta, 2\sin \theta)\) on this bigger circle. \(theta\) is the angle of rotation, counterclockwise, from the rightmost point on the circle.
Like any other circle, moving \(240\) degrees clockwise is the same as moving \(120\) degrees counterclockwise, so \(\theta = 120^{\circ}\). Thus our desired coordinate is
$$(2\cos 120^{\circ}, 2\sin 120^{\circ})$$
We just need to simplify. Even though the wheel is not a unit circle, we can still use the unit circle to recall the values of the trigonometric expressions. We get
$$\cos 120^{\circ} = -\frac{1}{2}$$
$$\sin 120^{\circ} = \frac{\sqrt{3}}{2}$$
Thus our coordinate is
$$\left(2 \cdot -\frac{1}{2}, 2 \cdot \frac{\sqrt{3}}{2}) \Rightarrow (-1, \sqrt{3}\right)$$
In general, if the radius of the circle is \(r\), the origin is \((x, y)\), and the degrees moved around the circle counterclockwise from the rightmost point on the circle is \(\theta\), then the point on the circle we are looking at is
$$(r\cos \theta + y, r\sin \theta + x)$$
It is easier to understand what this formula means than it is to memorize it. Understand where each part of the expression comes from, and you'll be in good shape.
From Example 1, note that the radius was \(2\), and origin is \(0, 0\), so we get the much simpler expression we used earlier; namely
$$(2\cos \theta, 2\sin \theta)$$
Example 2: A wheel of radius \(\frac{2}{3}\) is placed at \((2, 3)\) on the Cartesian Plane. We place a red dot at the rightmost point on the wheel and spin it \(570\) degrees counterclockwise. At what point on the Cartesian Plane is the red dot now at?
Solution: Let's work with the rotation first. As \(360\) degrees is a full revolution around the circle, rotating the wheel \(570\) degrees counterclockwise is the same as rotating the wheel \(210\) degrees counterclockwise. The dot will be in the same position regardless. Thus we let \(\theta = 210^{\circ}\).
If this were a unit circle centered at the origin, we would have
$$(\cos 210^{\circ}, \sin 210^{\circ})$$
However, there are two modifications to the unit circle that are made for this wheel: the position of the origin, and the radius. Always adjust the radius first. It is \(\frac{2}{3}\), so we multiply each coordinate by \(\frac{2}{3}\). Now we have
$$\left(\frac{2}{3}\cos 210^{\circ},\frac{2}{3} \sin 210^{\circ}\right)$$
Now we must shift the wheel so that its center is now \((2, 3)\). We translate the x-coordinate to the right by \(2\) units and the y-coordinate upward by \(3\) units.
$$\left(\frac{2}{3}\cos 210^{\circ} + 2, \frac{2}{3}\sin 210^{\circ} + 3\right)$$
This is the final coordinate pair. Simplify, using the Unit Circle for the trigonometric expressions within:
$$\left(\frac{2}{3} \cdot -\frac{\sqrt{3}}{2} + 2, \frac{2}{3} \cdot -\frac{1}{2} + 3\right) \Rightarrow \left(-\frac{\sqrt{3}}{3} + 2, -\frac{1}{3} + 3\right) \Rightarrow \left(\frac{6 - \sqrt{3}}{3}, \frac{8}{3}\right)$$