Distance to a Quadratic bezier curve

Recently, after viewing this illuminating animated GIFs, I realized that finding the closest point on a quadratic Bezier curve to a given point M might be not so difficult ...and very useful for detecting collision between a disk and a quadratic curve.

Quadratic Bezier curves are familiar to flash programmers, because these are the curves they draw with functions like Movieclip.moveTo() and Movieclip.curveTo().

Such Bezier curve is defined by 3 points: P0, P1, P2.
P0 and P1 are the extreme points, and P1 is a middle point determining curvature.

The curve can be parametrized with the formula:

P(t) = (1-t)²P0 + 2t(1-t)P1 +t²P2.

t is a parameter variable walking from 0 to 1. When t = 0, you are at point P0, and when t = 1, u're at P2.Each point of the curve has a corresponding "t".
The above-mentioned animated GIFs greatly help to "geometrically" grasp not only the quadratic curves, but also higher order Bezier curves .

Now, the key step is to get the derivative (back to school!! ) of P(t):

dP/dt(t) = -2(1-t)P0 + 2(1-2t)P1 + 2tP2

which can be rewritten as:

dP/dt(t) = 2(A+Bt)
where A = (P1-P0) and B = (P2-P1-A)

(note: A and B have a geometric signification: they are the diagonals of the parallelogram defined by P0,P1,P2 and a fourth point kind of "opposite" from P1)

The derivative is a vector representing the "speed" at point defined by P(t). It also has the property to be a tangent to the curve at point P(t).

Now let M(x,y) be our point (which can be anywhere) and we want to find out a point P' (defined by t') on Bezier curve that is the closest possible to M.

P(t') will meet the condition:

MP.dP/dt = 0
(where "." is the dot product of 2 vectors)

that is, MP will be orthogonal to the tangent at P.
Rewritting this condition yields:

(M - (1-t)²P0 + 2t(1-t)P1 +t²P2).(A+Bt) = 0

...After a (boring ) calculus you get a third degree equation (ouch!! ):

at3 + bt² + ct + d = 0,

where a = B², b = 3A.B, c = 2A²+M'.B, d = M'.A, (and M' = P0-M)

If we can solve this equation, we get 3 possible solutions for t !!
How to solve a third degree equation is explained here (in french).
Now we have a maximum of 3 solutions t0, t1, t2 and find the one fitting in [0,1] and minimizing dist(M, P(t)) is easy !!

Here is a demo:

Drag the circled dots
P0 and P2 are the blue points, P1 is green , M is red.
The yellow dot is the closest point, the orange dot indicates the normal to M.

After this, I realized that Andre Michelle had already posted a ball collision detection on his blog a long time ago. But the source wasn't available (at least I didn't work for nothing )...

The source code (AS2, FLA CS3 + Bezier.as).
Translating it into as3 should be no problem .