I'm not mathematically inclined, so please be patient with my question.


  • $(x_0, y_0)$ and $(x_1, y_1)$ as the endpoints of a cubic Bezier curve.

  • $(c_x, c_y)$ and r as the centerpoint and the radius of a circle.

  • $(x_0, y_0)$ and $(x_1, y_1)$ are on the circle.

  • if it makes the calculation simpler, it's safe to assume the arc is less than or equal to $\frac{\pi}{2}$.

How do I calculate the two control points of the Bezier curve that best fits the arc of the circle from $(x_0, y_0)$ to $(x_1, y_1)$?


3 Answers 3


Let $(x,y)^R=(-y,x)$ represent rotation by $\pi/2$ counterclockwise and $$ \gamma(t)=(1-t)^3p_0+3t(1-t)^2p_1+3t^2(1-t)p_2+t^3p_3 $$ define a cubic bezier with control points $\{p_0,p_1,p_2,p_3\}$.

Suppose $p_0=(x_0,y_0)$, $p_3=(x_1,y_1)$, and $c=(c_x,c_y)$ are given so that $|p_0-c|=|p_3-c|=r$ ($p_3$ is counterclockwise from $p_0$). Then $p_1=p_0+\alpha(p_0-c)^R$ and $p_2=p_3-\alpha(p_3-c)^R$ where $$ \alpha=\frac43\tan\left(\frac14\cos^{-1}\left(\frac{(p_0-c)\cdot(p_3-c)}{r^2}\right)\right) $$ For a quarter of a circle, $\alpha=\frac43(\sqrt2-1)$, and $\gamma$ is no more than $0.00027$ of the radius of the circle off.

Here is a plot of $\gamma$ in red over the quarter circle in black. We really don't see the circle since it is no more than $0.1$ pixels off from $\gamma$ when the radius is $400$ pixels.

$\hspace{3.5cm}$enter image description here

Computation of $\boldsymbol{\alpha}$

Looking at an arc with an angle of $\theta=\cos^{-1}\left(\frac{(p_0-c)\cdot(p_3-c)}{r^2}\right)$

$\hspace{1.5cm}$enter image description here

we see that the distance from $c$ to the middle of the arc is $$ r\cos(\theta/2)+\frac34\alpha r\sin(\theta/2) $$ we wish to choose $\alpha$ so that this is equal to $r$. Solving for $\alpha$ gives $$ \begin{align} \alpha &=\frac43\frac{1-\cos(\theta/2)}{\sin(\theta/2)}\\ &=\frac43\tan(\theta/4) \end{align} $$

A Slight Improvement

Using a circle of radius $1$, the maximum error in radius produced using $\alpha=\frac43\tan(\theta/4)$ is approximately $$ 0.0741\cos^4(\theta/4)\tan^6(\theta/4) $$ and the error is always positive; that is, the cubic spline never passes inside the circle. Reducing $\alpha$ reduces the midpoint distance by $\frac34\sin(\theta/2)=\frac32\tan(\theta/4)\cos^2(\theta/4)$ times as much, so to distribute the error evenly between the positive and negative, a first guess, assuming that the amplitude of the radius is unchanged, would be to reduce $\alpha$ by $0.0247\cos^2(\theta/4)\tan^5(\theta/4)$.

A bit of investigation shows that, when equalizing the positive and negative swings of the radius, the amplitude increases and that $$ \alpha=\frac43\tan(\theta/4)-0.03552442\cos^2(\theta/4)\tan^5(\theta/4) $$ gives pretty even distribution of the error between positive and negative for $\theta\le\pi/2$. The maximum error, both positive and negative, is approximately $$ 0.0533\cos^4(\theta/4)\tan^6(\theta/4) $$ When $\theta=\pi/2$, this agrees with the article mentioned by bubba in comments.

Note however, that in minimizing the radial error from the circle, the actual variation in radius is increased. Using the simple formula for $\alpha$, which puts the cubic bezier outside the circle, the radius varies by $0.0741\cos^4(\theta/4)\tan^6(\theta/4)$. However, when we minimize the error, the radial variation increases to $0.1066\cos^4(\theta/4)\tan^6(\theta/4)$.

  • 1
    $\begingroup$ This is a very good approximation of a circular arc, and it's the one that's most often used. However, it produces a curve that is everywhere outside the true circular arc. You can get a slightly better approximation by reducing the length of $\alpha$. In fact, the error for a unit radius quarter circle is reduced from 0.00027 to 0.00019. The details are here: spencermortensen.com/articles/bezier-circle $\endgroup$
    – bubba
    Commented Jul 22, 2014 at 6:09
  • 2
    $\begingroup$ Would the downvoter care to comment? $\endgroup$
    – robjohn
    Commented Jul 22, 2014 at 19:15
  • 1
    $\begingroup$ Nice. I agree that the accuracy of the original approximation would be fine for almost all applications. To get better approximations, you can apply the same sort of equi-oscillatory construction with higher degree polynomials (though it gets much harder). The accepted answer to this question has the details: math.stackexchange.com/questions/271319/… $\endgroup$
    – bubba
    Commented Jul 23, 2014 at 11:03
  • 1
    $\begingroup$ In the next comment, I'll post a Sage script that creates a plot comparing the vertical error (rather than the radial error) for a quarter circle arc, for various values of alpha: 'mid': the Bézier curve touches the midpoint of the arc, 'mort': Mortensen's value that minimizes radial drift, 'area': the curve and arc have equal area, from me22's answer. (You can test additional values by appending them to the params dictionary). $\endgroup$
    – PM 2Ring
    Commented Jun 28, 2021 at 10:16
  • 2
    $\begingroup$ If you look at Bézier Curve, the first diagram is of a cubic bezier curve. The curve is given by $$\gamma(t)=P_0(1-t)^3+3P_1t(1-t)^2+3P_2t^2(1-t)+P_3t^3$$ where the $P_k$ are the control points. $P_0$ and $P_3$ are the extreme control points and $P_1$ and $P_2$ are the interior control points. $\endgroup$
    – robjohn
    Commented Feb 16, 2023 at 20:05

As a different way to choose where the error falls, one could pick the control points such that the enclosed area matches that of the circle.

If we consider the first quarter of a circle (the same as the first image in the other answer), then the area should be $\pi/4$, so we need the integral

$$ \begin{align} \pi/4 &= \int_{0}^{1} y(t)x'(t)dt \\ &= \int_{0}^{1} ( (1-t)^3 + 3(1-t)^2t + 3(1-t)t^2c )( 3(1-t)^2c + 6(1-t)t(1-c) ) dt \\ &= -\frac{3c^2}{20}+\frac{3c}{5} + \frac12 \\ c &= 2 - \sqrt{\frac13(22-5\pi)} \\ &≈ 0.5517784778... \end{align} $$

That's slightly smaller than the other two answers, as it attempts to even out the error between over- and under-shooting the circle, rather than minimizing the maximum error.

Aside: Edits appreciated to add a solution for angles other than 90°. My maths are rusty.

  • 1
    $\begingroup$ You may like to look at the plot script I posted in a comment on robjohn's answer. $\endgroup$
    – PM 2Ring
    Commented Jun 28, 2021 at 10:19
  • $\begingroup$ Your answer is the closest to the 0.551784777779014 that a class in the .NET Framework uses for what they call the "ideal ratio of cubic Bezier points for a quarter circle." Many other repos on GitHub also use that magic value. $\endgroup$ Commented Nov 6, 2023 at 19:54
  • 1
    $\begingroup$ @LouisWaweru Tracing the git history to the PR, <github.com/dotnet/Microsoft.Maui.Graphics/pull/…> says that number came from <pomax.github.io/bezierinfo/#circles_cubic>, which appears to get it from a python binary search that tries to minimize the average error for t ∈ {⅕, ⅖, ⅗, ⅘}. I'm not sure why just those, but the page calls the improvements "essentially insignificant", so I don't think it was too concerned about getting really the best answer, just something that doesn't always overshoot when the constant can be hard-coded. $\endgroup$
    – me22
    Commented Nov 8, 2023 at 4:02

I got inspired to attempt yet another way to interpret "best fit": most-constrained curvature.

Let's look at the first quarter-circle, using the four control points $<1,0>$, $<1,α>$, $<α,1>$, $<1,0>$.

$$ \begin{align} x(t) &= (1-t)^3 + 3 (1-t)^2 t + 3 α (1-t) t^2 \\ y(t) &= 3 α (1-t)^2 t + 3 (1-t) t^2 + t^3 \\ &= x(1-t) \\ k(t) &= \frac{ x' y'' - y' x'' }{ (x'^2 + y'^2)^{3/2} } \\ &= \frac{ -18(2α(t - 1)t + αt^2 - 2(t - 1)t)(2α(t - 1) + αt - 2t + 1) + 18(α(t - 1)^2 + 2α(t - 1)t - 2(t - 1)t)(α(t - 1) + 2αt - 2t + 1) } { 27((α(t - 1)^2 + 2α(t - 1)t - 2(t - 1)t)^2 + (2α(t - 1)t + αt^2 - 2(t - 1)t)^2)^{3/2} } \end{align} $$

Where $k(t)$ is the signed curvature, but since we're going clockwise it'll be positive.

As is usually the case, the curvature expression is horrible, so we need a more tractable way forward.

Thinking about the extreme cases, a lower value of $α$ (like $0.1$) will make the ends turn sharply to get a flat middle while a high value of $α$ (like $1$) will make flat ends thus causing a pointy middle.

Graphing the curvature for a few values of $t$ helps confirm that intuition (source) curvature for various values of t

($t > 1/2$ not included on the chart due to symmetry.)

Since we're approximating a unit circle we're hoping to get $k(t) ≈ 1$, and those lines seem to intersect around there, at roughly $t ≈ 0.55$.

To balance out the extremes, we can pick $α$ such that ends and the middle have the same curvature. To avoid weird curves we'll require $α \in (0, 1)$, which simplifies a few things

$$ \begin{align} k(0) = k(1) &= k(1/2) \\ \frac{-2(α-1)}{3α^2} &= \frac{ 8\sqrt{2} α }{ 3(α-2)^2 } \\ {-2 (α-1) }{ 3(α-2)^2 } &= { 8\sqrt{2} α }{3α^2} \\ α &≈ 0.550581172753306 \end{align} $$

That value gives curvature within about $±1\%$ for $t \in [0, 1]$ (source) enter image description here

Exercise for the reader: Convince a CAS to compute

$$ \int_0^1 (k(t) - 1)^2 \mathrm{d}t $$


$$ \int_0^1 log(k(t))^2 \mathrm{d}t $$

and find the value of $α$ that actually minimizes the error in curvature -- it probably gives a slightly different value from the one I computed here.


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