$B$ doesn't appear in the problem so lets forget about $B$. Lets say that for any point $P$, $O_P$ is the circle at $P$ of some fixed radius $r$. E.g. $O_A$ is the circle at $A$, $O_C$ is the circle at $C$. I'll assume that the situation is set up so that what you're asking for is possible, so I won't try too hard to figure constraints on the problem.
Now say there's a point $C'$ on the line $CD$ such that the two circles $O_A$ and $O_{C'}$ kiss; going any further along $CD$ means the two circles will separate.
Exactly at the point $C'$, the distance between $A$ and $C'$ has to be $2r$. Algebraically this is
$$ \| A - C' \|^2 = 4r^2$$
where $\|V\|=\sqrt{V_1^2 + V_2^2}$ is the Euclidean norm of a vector. We will also use the dot product of two vectors $V\cdot W := V_1W_1 + V_2W_2$. Note that $\|V\|^2=V\cdot V$.
To determine this point $C'$, we will also need to know that $C'$ lies on the line $\vec{CD}$, which gives a formula if you know the coordinates of $C,D$: there is some $t\in[0,1]$ such that
$$ C' = C + t(D-C)$$
So to find $C'$, we just need to find this $t$. Plug this into the above:
$$ \|A-C - t(D-C)\|^2 = 4r^2$$
this is just a quadratic equation in $t$:
$$ \|A-C\|^2 -4r^2- 2t (A-C)\cdot(D-C) + t^2\|D-C\|^2 = 0$$
For
$$ a = \|D-C\|^2 , \quad b = -2 (A-C)\cdot(D-C), \quad c = \|A-C\|^2 -4r^2$$
the quadratic formula $t = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ gives
$$t = \frac{(A-C)\cdot(D-C) \pm \left[((A-C)\cdot(D-C))^2-\|D-C\|^2( \|A-C\|^2 -4r^2)\right]^{1/2}}{\|D-C\|^2} $$
Note that $\|A-C\|<2r$ (this is the assumption that the two circles $O_A,O_C$ intersect), so since we want $t\in[0,1]$, in particular $t\ge 0$ so we should choose
$$t = \frac{(A-C)\cdot(D-C) + \left[((A-C)\cdot(D-C))^2-\|D-C\|^2( \|A-C\|^2 -4r^2)\right]^{1/2}}{\|D-C\|^2} $$
Or equivalently
$$t = \frac{ \left[((A-C)\cdot(D-C))^2+\|D-C\|^2( 4r^2 +\|A-C\|^2 )\right]^{1/2}+(A-C)\cdot(D-C) }{\|D-C\|^2 }$$
If $AC$ and $DC$ were orthogonal, then $(A-C)\cdot(D-C) = 0$. Then $t$ simplifies to
$$ t=\frac{\sqrt{4r^2 -\|A-C\|^2 }}{\|D-C\| }$$
If $(A-C)\cdot(D-C) $ was small, you could try a linear approximation
$$ t \approx \frac{\sqrt{4r^2 -\|A-C\|^2 }}{\|D-C\| } + \frac{(A-C)\cdot(D-C)}{\|D-C\|^2} $$
In the general case, we have
$$C' = C + \frac{ \left[((A-C)\cdot(D-C))^2+\|D-C\|^2( 4r^2 -\|A-C\|^2 )\right]^{1/2}+(A-C)\cdot(D-C) }{\|D-C\|^2 }(D-C)$$