Why it is impossible to draw a circle by Bezier Curve?

Question

I have see a lot of people said it is not possible to draw a circle by Bezier Curve.

However, I want to know why?

I did see somebody explain, but I am not quite sure what they mean.

Like, Bezier curve is parabola, so it is impossible to draw a circle by it.

But how can I proof it? proof that Bezier curve is only parabola and never be circle?

Answer 1

Hint:

Consider that the coordinates of a Bezier curve are polynomials in $t$, and that the coordinates of a circle all lie on $x^2+y^2=1$ (without a loss of generality).

What can we say about polynomials $x(t),y(t)$ such that $x(t)^2+y(t)^2=1$?

Answer 2

A Bezier curve can be quadratic or cubic or even have a higher degree.

It seems you are concerned only by quadratic Bezier curves defined by distinct points $A_1(x_1,y_1)$ and $A_3(x_3,y_3)$, and controlled by point $A_2(x_2,y_2)$ with equations:

$$\binom{x}{y}=\binom{x_1}{y_1}(1-t)^2+\binom{x_2}{y_2}2t(1-t)+\binom{x_3}{y_3}t^2 \ \ \text{for} \ \ 0 \leq t \leq 1,$$

$$\tag{0} \Longleftrightarrow \begin{cases}{x=at^2+bt+c\\y=dt^2+et+f.}\end{cases}$$

for certain real numbers $a,b,c,d,e,f$.

Were this curve an arc of circle, one would have, for a certain center $(x_0,y_0)$ and radius $R$:

$$(x-x_0)^2+(y-y_0)^2=R^2$$

i.e., taking (0) into acount:

$$\tag{1} \forall t \in (0,1): (at^2+bt+c-x_0)^2+(dt^2+et+f-y_0)^2-R^2=0$$

Expanding the LHS expression and expressing that all the coefficients of $t^1, t^2, t^3, t^4$ are zero :

$$\tag{2} \begin{cases}bc + ef = 0 \\ b^2 + e^2 + 2ac + 2df = 0 \\ ab + de = 0 \\ a^2 + d^2 = 0 \end{cases}$$

it is immediate to deduce (using the 4th equation) that $a=d=0$ and then plugging these values into the second equation, that $b=e=0$. Thus looking at (1), we see that our Bezier curve should be reduced to a point, bringing a contradiction.

Added remark 1: (0) represents in fact a parabolic arc, because it is the affine image of the "archetypal" parabola $X=Y^2$ because $(X=t^2;Y=t):$

$$\pmatrix{x\\y}=\pmatrix{a&b\\d&e}\pmatrix{t^2\\t}+\pmatrix{c\\f}$$

being known that the affine image of a conic is a conic with the same type.

Added remark 2: If one accepts rational fractions instead of polynomials, i.e., any circle can be obtained as a so-called rational Bézier curve ; for example the unit circle is given by the parametric equations:

$$\begin{cases}x=\dfrac{1-t^2}{1+t^2}\\y=\dfrac{2t}{1+t^2}\end{cases}$$

(for funny figures and comments, see (https://maxwelldemon.com/2014/03/03/rational-parameterisation-of-the-circle/))