The standard formula for a cirlce is $\displaystyle x^2+y^2=r^2$. This can also be written as $\displaystyle x^2+y^2-r^2=0$ or more properly as $\displaystyle C(x,y)=x^2+y^2-r^2=0$.

There are situations where the parametric form of the circle is useful.

Here, I will show how to derive the parametric form of the circle using angles on the circle and a parameter $\displaystyle t$.

Given a circle $\displaystyle C$, with radius $r=1$;

we can derive its parametric form in 2 ways;

  1. Representing the point B on the circle in terms of the line $L$.

    the equation for line $\displaystyle L$ is given by $\displaystyle y=tx-t=t(x+1)$ or $ L(x,y) = (x,\ t(x+1))$. We can then represent the circle $\displaystyle C$ in terms of $\displaystyle L$ as $\displaystyle f(x) = C(L) = C(x,\ t(x+1))=x^2+t^2(x+1)^2-1^2=0$. Expanding $\displaystyle f(x)$, we end-up with the quadratic equation $\displaystyle f(x) = x^2 + \frac{2xt^2}{1+t^2} + \frac{t^2-1}{1+t^2} = 0$. Solving for $\displaystyle x$ and $\displaystyle y$ we get:

    $$B(x,y)=\Big (\frac{1-t^2}{1+t^2},\ \frac{2t}{1+t^2} \Big) \qquad -1 \le t \le 1$$

    .

    Note:

    We can solve the quadratic equation with;

    1. quadratic formula, or
    2. remainder theorem.

    Since we already know that $ x=-1$ is a solution, that is $ f(-1)=0$ then $x-(-1)=x+1$ is a factor. We can avoid using the quadratic formula and instead use the remainder theorem, to solve for the second solution for $ x$.

  2. Representing the point B on the circle in terms of angles $\theta$ and $\phi$.

    From the diagram, we see that $\displaystyle B(x, y)=(r*\cos(\phi),\ r*\sin(\phi))$. Using Thale’s Theorem, the angle $\displaystyle \phi$ with respect to angle $\displaystyle \theta$ is $\displaystyle \phi=2*\theta$. Therefore, $\displaystyle B(x, y)=(r*\cos(2*\theta),\ r*\sin(2*\theta))=(\cos(2*\theta),\ \sin(2*\theta))$.

    $$\cos(2*\theta)=2*\cos^2(\theta) - 1 \quad$$$$\sin(2*\theta)=2*\sin(\theta)*\cos(\theta)$$$$B(x, y)=(2*\cos^2(\theta) - 1,\ 2*\sin(\theta)*\cos(\theta))$$

    From the diagram, the hypotenuse $h$, the line segment $TA$, is $\displaystyle h^2=r^2 + t^2=1+t^2$. Hence, $\displaystyle \cos(\theta)=1/h$ and $\displaystyle \sin(\theta)=t/h$.

    $$B(x,y)=(2*\cos^2(\theta) - 1,\ 2*\sin(\theta)*\cos(\theta))$$$$=\Big(2*\Big(\frac{1}{h}\Big)^2 - 1,\ 2*\frac{t}{h}*\frac{1}{h}\Big)$$$$=\Big(\frac{2-h^2}{h^2},\ \frac{2*t}{h^2}\Big) \qquad \qquad \quad $$$$\qquad \qquad =\Big(\frac{1-t^2}{1+t^2},\ \frac{2t}{1+t^2}\Big) \qquad \quad -1 \le t \le 1 $$

Half angle formulae

From this result we note that;

$$B(x, y)=(\cos(\phi),\ \sin(\phi))$$$$\cos(\phi)=\frac{1-t^2}{1+t^2}$$$$\sin(\phi)=\frac{2t}{1+t^2}$$$$\tan(\phi)=\frac{\sin(\phi)}{\cos(\phi)}=\frac{2t}{1-t^2}$$

the last result is known as the half-angle formulae which is important in solving certain class of calculus problems.