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
x^{2}+y^{2}=1 means that y=(1-x^{2})^{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-x^{2})^{1/2}

t = (1-x^{2})^{1/2}/(x+1)

t(x+1) = (1-x^{2})^{1/2}

t^{2}(x+1)^{2} = 1-x^{2}

t^{2}(x^{2}+2x+1) = 1-x^{2}

(t^{2}+1)x^{2}+2t^{2}x+t^{2}-1=0

x = (1-t^{2})/(1+t^{2})

Plugging this expression in
for x in the formula x^{2}+y^{2}=1 we get

y = 2t/(1+t^{2})

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

x(t) = (1-t^{2})/(1+t^{2})

y(t) = 2t/(1+t^{2})

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

x^{2}+y^{2} = ((1-t^{2})/(t^{2}+1))^{2}+(2t/(t^{2}+1))^{2}
= (t^{4}-2t^{2}+1)/((t^{2}+1)^{2})+4t^{2}/)/((t^{2}+1)^{2})
= (t^{4}+2t^{2}+1)/ ((t^{2}+1)^{2}) = (t^{2}+1)^{2}/(t^{2}+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!