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The definition of a circle is simple: it is a set of points located at the same distance from a given point. This distance is called the radius and this point is called the centre of the circle. The circle with centre $$(-1,-3/2)$$ and radius $$\sqrt 6$$
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## 1. Circles as sets of solutions of an equation

But what is a point, and what is a distance? If we have a very clear intuition, this one is not sufficient to speak in a rigorous way about these things in mathematics.

We mentioned in the article The Euclidean Plane: Ancient Geometry and the Analytical Approach how it is possible to give a precise meaning to the notion of a point and the notion of a distance in the framework of the modern Euclidean plane. Thanks to the modern construction of real numbers, every positive real number has a square root, and two points $$M$$ and $$N$$ being represented by their respective coordinates $$(x,y)$$ and $$(t,u)$$ in the Cartesian approach to geometry, the distance $$d(M,N)$$ between these two points is the square root of the sum of the squares of the distances between their coordinates, i.e. $$d(M,N)=\sqrt{(x-t)^2+(y-u)^2}$$, which is a reformulation of the Pythagorean theorem.

From this point of view, we have a perfectly clear “analytical” description of what a circle is in the plane, from the definition at the start: the circle $$C$$ with centre $$I=(a,b)$$ and radius $$R$$ (positive real number) is the set of points $$M=(x, y)$$ of the plane $$\mathbb R^2$$ such that the distance $$d(I,M)$$ between $$I$$ and $$M$$ is $$R$$, that is, since the distance $$d(I,M)$$ is $$\sqrt{(x-a)^2+(y-b)^2}$$, symbolically $\mathscr C=\{(x,y)\in\mathbb R^2 | \sqrt{(x-a)^2+(y-b)}=R\}.$ As the number $$R$$ is positive, we can then do without the square root to finally describe the circle $$C$$ as the set $\mathscr C=\{(x,y)\in \mathbb R^2 | (x-a)^2+(y-b)^2=R^2\},$ \which gives us the typical equation of a circle in the plane, here $$(x-a)^2+(y-b)^2=R^2$$ for the circle $$C$$ with centre $$I=(a,b)$$ and radius $$R$$.

In other words, such a circle is the set of solutions $$(x,y)$$ of this equation: thanks to Descartes’ analytical method, we can describe a circle in the plane as a set of solutions of an equation, in a way analogous to the description of a line as a set of solutions of an equation. For example, the circle at the beginning has equation $$(x+1)^2+(y+\frac 3 2)^2=6$$.

## 2. The trigonometric circle : Cosine and Sine as angular coordinates

Now, we have at our disposal by means of real analysis (the theory of functions of the set $$\mathbb R$$) two functions, cosine and sine, known as “trigonometric”, and which give us the coordinates of a point of the trigonometric circle, which is the circle with centre $$O=(0,0)$$ and radius $$1$$. Indeed, a point $$M=(x,y)$$ on this circle is marked by the angle it describes with the axis $$[0x)$$, which is measured as the (oriented) length of the circular arc bounded by the point $$I=(0,1)$$ and the point $$M$$, which is called the measure in radians of the angle $$\widehat{IOM}$$. In other words, if the length of the arc $$\overset{\frown}{IM}$$ is $$t$$, the (Cartesian) coordinates of the point $$M$$ are $$x=\cos t$$ and $$y=\sin t$$ : the cosine and sine of the angle $$t$$ are the projections of the point $$M$$ on each axis. The trigonometric circle and the coordinates of a point on that circle, expressed as the cosine and sine of the angle $$\widehat{IOM}$$ with measure $$t$$ radians.

Since the radius of the trigonometric circle is equal to $$1$$, this means by the Pythagorean theorem that, the angle determined by the point $$M$$ being equal to $$t$$, we have the equality $$cos^2 t+sin^2t=1$$. In general, for any real number $$t$$ this equality is true and $$\cos t$$ and $$\sin t$$ are the coordinates of a point of the trigonometric circle: in other words, the couples of real numbers $$(x,y)$$ of the form $$x=cos t$$ and $$y=sin t$$ verify the equation $$x^2+y^2=1$$, which is the equation of the trigonometric circle, since it can be rewritten as $$(x-0)^2+(y-0)^2=1^2$$, which, as we have seen, is the equation of the circle with centre $$(0,0)$$ and radius $$1$$. Conversely, any point $$(x,y)$$ verifying this equation, is on the trigonometric circle, and is of the form $$\cos t,\sin t$$ as we have discussed, for a real number $$t$$ taken in the interval $$[0,2\pi[=\{x(t)\in \mathbb R | 0\leq t<2\pi\}$$.

## 3. Parameterising a circle: going from an equation to a plot

We can therefore adopt another approach to describe the trigonometric circle, i.e. describe it by a parameter, which is mathematically equivalent to “plotting” it in the plane, using a function $$f:[0,2\pi[\to \mathbb R^2$$ : a function is a mathematical operation that “transforms” one object into another; here, the transformed object is the parameter $$t$$ (a real number greater or equal to $$0$$ and lesser than $$2\pi$$), and the result of the transformation is the point $$f(t)=(\cos t,\sin t)$$ of the plane, which describes the “value of the function $$f$$ at point $$t$$”. When $$t$$ takes all the values between $$0$$ and $$2\pi$$, then $$f(t)$$ takes as values all the points of the trigonometric circle; if we represent $$t$$ as “time”, we have conceptualised the drawing of the trigonometric circle, in contrast with its definition by an equation. We say that we have parameterised this circle, that the function $$f$$ is a parametrisation of the trigonometric circle.

Using this parametrisation, we can then describe a parametrisation, a “plot”, of any circle in the plane – at least if its radius is not zero – by returning to the description by an equation. If, as before, $$\mathscr C$$ is the circle with centre $$I=(a,b)$$ and radius $$R>0$$, whose equation is $$(x-a)^2+(y-b)^2=R^2$$, a point $$M=(x,y)$$ is on $$\mathscr C$$ if and only if the preceding equation is verified. Now, this equation can be rewritten in the following form: dividing by $$R^2$$ on each side, it is equivalent to the equation $\left(\dfrac{x-a}{R}\right)^2+\left(\dfrac{y-b}{R}\right)^2=1,$ so that the point $$M=(x,y)$$ is on the circle $$\mathscr C$$ if and only if the point $$(\frac{x-a}{R},\frac{y-b}{R})$$ is on the trigonometric circle! This is equivalent to having $$x=a+R\cos t$$ and $$y=a+R\sin t$$, so we can “draw” the circle $$\mathscr C$$ thanks to the parametrisation $$g:[0,2\pi[\to \mathbb R^2$$ defined by $g(t)=(a+R\cos t,a+R\sin t).$ One could replace the interval $$[0,2\pi[$$ by another interval, or change the “speed” of the plot, but the main point here is to remember that a circle in the plane can be defined either by an equation or by a parameter, and that one can switch from one description to the other. These are the two fundamental ways of describing a geometric object.