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.
$\hspace{3.5cm}$
Computation of $\boldsymbol{\alpha}$
Looking at an arc with an angle of $\theta=\cos^{-1}\left(\frac{(p_0-c)\cdot(p_3-c)}{r^2}\right)$
$\hspace{1.5cm}$
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 $0.1066\cos^4(\theta/4)\tan^6(\theta/4)$.
The simplest approach is to use $N+1$ points to construct a curve of degree $N$. You have to assign a parameter ($t$) value to each point. So, to construct a cubic curve through four given points $\mathbf{P}_0$, $\mathbf{P}_1$, $\mathbf{P}_2$, $\mathbf{P}_3$, you need four parameter values, $t_0, t_1, t_2, t_3$. Then, as @fang said in his answer, you can construct a set of four linear equations and solve for the four control points of the curve. Two of the equations are trivial, so actually you only have to solve two equations.
The simplest approach is to just set
$$t_0 = 0 \quad , \quad
t_1 = \tfrac13 \quad , \quad
t_2 = \tfrac23 \quad , \quad
t_3 = 1$$
Then the matrix in the system of linear equations is fixed, and you can just invert it once, symbolically. You can get explicit formulae for the control points, as given in this question. But this only works if the given points $\mathbf{P}_0$, $\mathbf{P}_1$, $\mathbf{P}_2$, $\mathbf{P}_3$ are spaced fairly evenly.
To deal with points whose spacing is highly uneven, the usual approach is to use chord-lengths to calculate parameter values. So, you set
$$c_0 = d(\mathbf{P}_0, \mathbf{P}_1) \quad ; \quad
c_1 = d(\mathbf{P}_1, \mathbf{P}_2) \quad ; \quad
c_2 = d(\mathbf{P}_2, \mathbf{P}_3)$$
Then put $c = c_0+c_1+c_2$, and
$$t_0 = 0 \quad , \quad
t_1 = \frac{c_0}{c} \quad , \quad
t_2 = \frac{c_0+c_1}{c} \quad , \quad
t_3 = 1$$
Then, again, provided $t_0 < t_1 < t_2 < t_3$, you can set up a system of linear equations, and solve.
If you're willing to do quite a bit more work, you can actually construct a cubic curve passing through 6 points. Though, in this case, you can't specify the parameter values, of course. For details, see my answer to this question.
Best Answer
As Joriki says, you can't get the other control point with no other information.
However, if you know a point on the line you can work it out. The formula for a quadratic Bezier function taken from https://en.wikipedia.org/wiki/Bezier_curve is:
$P(t) = (1-t)^2P0 + 2(1-t)tP1 + t^2P2$
If you know any two control points, and a separate point on the line at a known time you can calculate the third control point. Say you know the start and end points P0 and P2, and know the curve passes through P(0.5) at t=0.5:
$$ P(0.5) = (1-0.5)^2P0 + 2(1-0.5)0.5P1+0.5^2P2\\ P(0.5) = 0.25P0 + 0.5P1 + 0.25P2\\ P1 = 2P(0.5)-0.5P0 - 0.5P2 $$
For a cubic Bezier if you know the start and end points you can find the middle two control points using the method described here: https://web.archive.org/web/20131225210855/http://people.sc.fsu.edu/~jburkardt/html/bezier_interpolation.html
That requires you to know the points that the curve goes through at t=1/3 and t=2/3. I'm afraid I haven't tried generalising it to any points in time.