Are the two red lines supposed to be perpendicular to one another, and at 45 degrees to the vertical, as shown? If so, and the circle center is at $(a, b)$ and the circle radius is $r$, then the points you're looking for are at
$$
(a + \frac{\sqrt{2}}{2} r, b + \frac{\sqrt{2}}{2}r) \\
(a - \frac{\sqrt{2}}{2} r, b + \frac{\sqrt{2}}{2}r).
$$
Since the two lines are (after the comments) not supposed to be perpendicular, let's call the angle between them $\alpha$; in your case, $\alpha = 85.702%{\circ}$. Let
$$\beta = 90^{\circ} - \frac{\alpha}{2};$$
Then $\beta$ is the half-angle between one of your dotted orange lines and the vertical greenish line. The offset of the two points to either side of that vertical line is therefore $r \sin \beta$, so we get
$$
(a + r \sin \beta, b + r \cos \beta) \\
(a - r \sin \beta, b + r \cos \beta).
$$
Applying these to your points, I get (using 3.14159 for PI, and Excel for the calculations)
right point = ( 0.073312432, -0.131990535)
left point = (-0.073312432, -0.131990535)
That's in pretty close agreement with your CAD results, and it's also mathematically correct. :)
Here is another approach that may be quicker and is accessible without vector calculus.
The two circles have equations $\left(x-h\right)^2+\left(y-k\right)^2=r^2$ and $\left(x-u\right)^2+\left(y-v\right)^2=s^2$
We'll just search for the point $\left(x_1,y_1\right)$ that is on the $\left(h,v\right)$-circle. Once we have that, it's easy to see that the slope of the tangent line is $\frac{y_1-k}{x_1-h}$. So the tangent line will have equation $$y=\frac{y_1-k}{x_1-h}\left(x-x_1\right)+y_1$$
Assume for a second that we have found $x_1$ and $y_1$. How will this line meet the second circle? We can make the substitution for $y$ into the second circle's equation:
$$\begin{align}
\left(x-u\right)^2+\left(y-v\right)^2&=s^2\\
\left(x-u\right)^2+\left(\frac{y_1-k}{x_1-h}\left(x-x_1\right)+y_1-v\right)^2&=s^2\\
\left(x-u\right)^2\left(x_1-h\right)^2+\left(\left(y_1-k\right)\left(x-x_1\right)+\left(y_1-v\right)\left(x_1-h\right)\right)^2&=s^2\left(x_1-h\right)^2\\
\end{align}$$
This is a quadratic equation in $x$, and $x$-values that solve it are $x$-values for a point on the second circle where the line will intersect it. We want this equation to have precisely one solution, since we want the line to be tangent to the second circle. So we rearrange the equation so that we can cleanly see the discriminant and set it equal to $0$.
$$\begin{align}
x^2\left(\left(x_1-h\right)^2+\left(y_1-k\right)^2\right)+x\left(-2u\left(x_1-h\right)^2-2x_1\left(y_1-k\right)^2+2\left(y_1-k\right)\right)\\
{}+u^2\left(x_1-h\right)^2+\left(y_1-k\right)^2x_1^2-2x_1\left(y_1-k\right)\left(y_1-v\right)\left(x_1-h\right)+\left(y_1-v\right)^2\left(x_1-h\right)^2\\
{}-s^2\left(x_1-h\right)^2&=0\\
\end{align}$$
And so along with satisfying $\left(x_1-h\right)^2+\left(y_1-k\right)^2=r^2$, $x_1$ and $y_1$ satisfy
$$\begin{align}
\left(B\right)^2-4\left(A\right)\left(C\right)&=0
\end{align}$$
where $A$, $B$, and $C$ are evident in the quadratic equation above. So that makes two equations in two unknowns $x_1$ and $y_1$. Solving these two equations will give you one point on the first circle, from which you can determine a slope and plot the tangent line.
Best Answer
As others have mentioned in comments, your control points cannot be independent. Nevertheless, if we assume that a given configuration has the properties you want, we can analyze the geometry.
I'll consider the orange arc, $BE$, and I'll assume that both circles $A$ and $C$ overlap the interior of the orange circle, which I'll further assume has not degenerated into a line.
Let $a = |AB|$, $c = |CE|$, and $x=|AC|$; all of these can be considered known quantities. Let the (unknown) radius of the orange circle be $r = |PB| = |PE|$, where $P$ is the circle's center. Because radii $AB$ and $PB$ are perpendicular to a common tangent line at $B$, these segments lie on the same line; likewise for $CE$ and $PE$; consequently, $P$ lies at the intersection of the two extended radii $AB$ and $CE$, so that the angle $BPE$ is congruent to the angle between the vectors $AB$ and $CE$. Call the measure of that angle $\theta$; it, too, can be considered a known quantity.
Now, triangle $APC$ has sides of length $x$, $r-a$, and $r-c$ (the last two because of the assumed overlap of circles), with angle $\theta$ between the last two. By the Law of Cosines:
$$x^2 = (r-a)^2 + (r-c)^2 - 2 (r-a)(r-c) \cos\theta$$
Solve this quadratic equation for $r$, and you can calculate whatever else you need to know: arc length, location of $P$, equation of the orange circle, etc.
(The equation of the orange circle can be expressed in a form that degenerates into a line equation as $r$ approaches infinity. Note that, in such a degenerate case, $\theta = 0$.)