I am trying to draw a circle using 4 Quadratic Bézier curves. By referring https://www.degruyter.com/document/doi/10.1515/math-2016-0012/html.
$$C(t) = (1-t)^2P_0 + 2(1-t)tP_1 + t^2P_2.$$
If the arc is like this, the $t = 0.5$ and $C(t) = Radius$, so I can get the an equation:
$$Radius = (1-0.5)^2cos(\theta)Radius + 2(1-0.5)0.5P_1 + t^2cos(\theta)Radius$$
then I can get the control point, and use these to draw a $1/4$ arc and rotate it to make a circle.
$$P_1 = 2Radius - cos(\theta)Radius$$
but the circle I drew looks like, is there anything I did wrong? How to make the circle round?
The Quadratic Bezier curve with control points $P_{0}$, $P_{1}$, and $P_{2}$ is initially tangent (i.e., at $P_{0}$) to the segment $P_{0}P_{1}$, and terminally tangent (i.e., at $P_{2}$) to $P_{1}P_{2}$. To get a "smooth" closed curve from four parabolas it's necessary to take the control points to be corners and side midpoints of a square:
This curve is also visibly not round. If you must use quadratic splines, the "optimal curve" may need to be determined qualitatively, balancing corners against over-extension along the axes.
Cubic splines can do better, but ultimately the problem is that a circle is not a union of splines.