Parametrizations of a Circle

By

Alex Moore

In this write-up we investigate parametrizations of circles. A parametric curve is defined as a collection of points given by two continuous functions x(t) and y(t), that is, the points on the curve are the collection of points (x(t), y(t)) where x and y are continuous functions of t. The way to think of parametric curves are as traced paths in the plane of a particle in space with t representing time. What we hope to obtain is a parametric interpretation of the circle.

Lets start with the easiest possible case. We wish to parametrize the unit circle. The unit circle is defined by the equation x^2 + y^2 =1. From elementary trigonometry we recall the identity (cos(t))^2 + (sin(t))^2 =1 for all [0, 2p). This directly gives us our first parametrization of the unit circle: Let x(t) = cos(t) and y(t) = sin(t).

This basic trigonometry identity is proof that this parametric curve is the unit circle.

This is only one example. What if we wanted to parametrize a circle of any given radius (assumed to be positive) centered at the origin? That is, we want to find a pair of functions x(t) and y(t) that give us the circle defined by x^2 + y^2 = a^2. This is the circle of radius a>=0. Consider the two functions x(t) = a cos(t) and y(t) = a sin(t). Then we calculate

a^2

and so we have that x^2 + y^2=a^2. Therefore, the equations x(t) = a cos(t) and y(t) = a sin(t) are parametric equations for our circle of radius a centered at the origin. Why does this simple modification work? Recall that -1<=cos(t), sin(t)<=1 for all values of t. by multiplying these functions by the constant a we extend (or shrink if a<1) the range to –a<=cos(t),sin(t)<=a. Since x(t)=a cos(t) the x-values of the graph now range from –a to a and since y(t)=a sin(t) the y-values range from –a to a. Now we take the final step in our generalization. We wish to adapt our current parametrization to any circle. That is, we want to parametrize the circle of radius a centered at (h,k), (x-h)^2 + (y-k)^2=a^2. Since we want to shift our circle centered at the origin to (h,k) we need to shift our x-coordinates horizontally by h and our y coordinates vertically by k. Since x(t) defines the x-coordinates the obvious guess would be to set x(t)=h+ a cos(t). Similarly, set y(t)=k+ a sin(t). Now we check:

x = h+ a cos(t), (x-h) = a cos(t), (x-h)^2=a^2(cos(t))^2

y = k + a sin(t), (y-k) = a sin(t), (y-k)^2 = a^2(sin(t))^2

Adding these equations together yield

(x-h)^2 + (y-k)^2 = a^2.

And we are successful in parametrizing a general circle! To demonstrate we do an example with a=3, h=1 and k= -2.

As mathematicians there is a natural question to ask next. Are parametrizations unique? That is, we want to know if the above parametrization of a circle the only one. The answer is no! Parametrizations are not unique. We will now find a second parametrization of a circle. We again reduce to the simple case of the unit circle and then extend to general circles in the plane. Consider the picture below:

Let t=m be the slope of a line passing through (-1,0). Notice that every line through (-1,0) intersects the circle at exactly one point. In this parametrization we get every point on the unit circle except (-1,0), which is the point that the line of infinite slope (vertical line x=-1) intersects the circle. Next we point out that since the distance from (-1,0) to the origin is 1, the line with slope t will have a y-intercept of (0,t). Recall that x²+y²=1 means that y=(1-x²)^1/2 (this could be positive or negative but we will be squaring this later so it does not matter). Now with an application of similar triangles and the quadratic formula we get that

1/t = (x+1)/(1-x²)^1/2

t = (1-x²)^1/2/(x+1)

t(x+1) = (1-x²)^1/2

t²(x+1)² = 1-x²

t²(x²+2x+1) = 1-x²

(t²+1)x²+2t²x+t²-1=0

x = (1-t²)/(1+t²)

Plugging this expression in for x in the formula x²+y²=1 we get

y = 2t/(1+t²)

To summarize, we have the parametric equation for the unit circle

x(t) = (1-t²)/(1+t²)

y(t) = 2t/(1+t²)

Let us quickly check that this is indeed the unit circle.

x²+y² = ((1-t²)/(t²+1))²+(2t/(t²+1))² = (t⁴-2t²+1)/((t²+1)²)+4t²/)/((t²+1)²) = (t⁴+2t²+1)/ ((t²+1)²) = (t²+1)²/(t²+1)² = 1

And so we indeed have our circle! The picture below has all values of t between -30 and 30. To get the whole circle (minus (-1,0)) we would need to take t to positive infinity and negative infinity.

To extend this parametrization of the unit circle to any circle in the plane is not difficult. In fact, it is exactly the same as what we did before. To parametrize the circle of radius r centered at (h,k) we simply set

And thus we now have two different parametrizations of the circle of radius r centered at (h,k). What other parametrizations are there? There are infinitely many more to find. Good luck!