Let $C$ denote the unit circle in the two-dimensional plane, centered at the origin (the blue circle in the pic). Let $E$ an ellipse whose equation is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. The set $C+E$, known as the Minkowski sum of $C$ and $E$, is defined by $$C+E=\{\vec{\alpha}+\vec{\beta}: \vec{\alpha}\in C,\ \ \vec{\beta}\in E\}$$ In the picture above, the orange oval-shaped enclosing figure is the boundary of the sum $C+E$, (where for the sake of this example, $E$ is the ellipse with $a=2,b=1$). The green ellipse in the picture is $Q+E$, where for the sake of example, $Q=(1/\sqrt{2},1/\sqrt{2})$. The point $P$ represents a typical boundary point of the sum $C+E$, which is the orange oval, which is the geometric location of all points obtained by rotating around the green ellipse with its center varying along the blue circle. The point $O$ is the origin $(0,0)$ and the point $M$ is the orthogonal projection of $P$ onto the $x$ axis.
Problem: Express the length of $\vec{OP}$ in terms of the angle POM.
I can prove that if $$F(\varphi)=\arctan\left(\tan\varphi \frac{\sqrt{a^2\cos^2\varphi+b^2\sin^2\varphi}+b^2}{\sqrt{a^2\cos^2\varphi+b^2\sin^2\varphi}+a^2}\right),\quad (0\leq\varphi\leq \pi/2)$$ Then if the angle POM is $\theta$, then the angle QOM is $F^{-1}(\theta)$ (the inverse function), and then it is not too difficult to express the $\vec{QP}$ in terms of $F^{-1}(\theta)$ and consequently the length of $\vec{OP}$ in terms of $\theta$ (and $a,b$ of course). However, the result seems to be too complicated; I suspect there might be a simpler geometric argument that escapes me.
HINT
Since Minkowsky sum is commutative, then the alternative order of the calculations can be proposed.
The point of the circle farthest from the ellipse is located on the normal to the ellipse.
Denote $$\angle xOQ = \varphi,\quad \angle xOP = \theta.$$
Parametric equation of the ellipse is $$x=a\cos\varphi,\quad y=b\sin\varphi,$$ the guide vector of the tangent line is $\{-a\sin\varphi,b\cos\varphi\},$ and the angle coefficient of the line $QP$ is $$k=\dfrac ab\tan\varphi.$$
Then \begin{align} &\overline{OQ}=\{a\cos\varphi,b\sin\varphi\}, \\[8pt] &\overline{QP}=\dfrac1{\sqrt{k^2+1}}\{1,k\} =\dfrac{\{b\cos\varphi,a\sin\varphi\}}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}},\\[8pt] &\overline{OP} = \overline{OQ}+\overline{QP}\\ &=\left\{\left(a+\dfrac{b}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)\cos\varphi, \left(b+\dfrac{a}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)\sin\varphi\right\}\\ &=\sqrt{\left(a+\dfrac{b}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)^2\cos^2\varphi + \left(b+\dfrac{a}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)^2\sin^2\varphi}\\ &\times\{\cos\theta,\sin\theta\},\\[8pt] &\left(a+\dfrac{b}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)^2\cos^2\varphi =\cos^2\theta\\ &\times\left(\left(a+\dfrac{b}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)^2\cos^2\varphi + \left(b+\dfrac{a}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)^2\sin^2\varphi\right),\\[8pt] &\left(a\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}+b\Large\mathstrut\right) \cos\varphi\sin\theta =\left(b\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}+a\right)\sin\varphi\cos\theta,\\[8pt] &(a\cos\varphi\tan\theta-b\sin\varphi)\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi} =a\sin\varphi - b\cos\varphi\tan\theta,\\[8pt] \end{align}
$$\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi} =\dfrac{a\sin\varphi - b\cos\varphi\tan\theta}{a\cos\varphi\tan\theta-b\sin\varphi},\tag1$$
$$(a\tan\theta-b\tan\varphi)^2(a^2\tan^2\varphi+b^2) = (a\tan\varphi - b\tan\theta)^2(1+\tan^2\varphi),\tag2$$ and fourth-order algebraic equation has known exact solution $$\tan\varphi = f(\tan\theta).$$
Also from $(1)$ should $$a+\dfrac{b}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}} = a+b\,\dfrac{a\cos\varphi\tan\theta-b\sin\varphi} {a\sin\varphi - b\cos\varphi\tan\theta} = (a^2-b^2)\dfrac{\sin\varphi}{a\sin\varphi - b\cos\varphi\tan\theta},$$ $$b+\dfrac{a}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}} = b+a\,\dfrac{a\cos\varphi\tan\theta-b\sin\varphi} {a\sin\varphi - b\cos\varphi\tan\theta} = (a^2-b^2)\dfrac{\cos\varphi\tan\theta}{a\sin\varphi - b\cos\varphi\tan\theta},$$ $$OP = (a^2-b^2)\dfrac{\sin\varphi\cos\varphi\sqrt{1+\tan^2\theta}}{a\sin\varphi - b\cos\varphi\tan\theta},$$ $$OP = (a^2-b^2)\dfrac{\sin\varphi}{a\tan\varphi\cos\theta - b\sin\theta}.\tag3$$
Formulas $(2),(3)$ define required result.
$\color{brown}{\mathbf{About\ equation\ (2).}}$
Substitutions $$t=a\tan\varphi-b\tan\theta,\quad \tan\theta = p\tag4$$ present equation $(2)$ in the form of $$(a^2p - b(t+bp))^2((t+bp)^2+b^2) = t^2(a^2+(t+bp)^2),$$ $$\Bigl(((a^2-b^2)p - bt)^2-t^2\Bigr)((t+bp)^2+b^2) = (a^2-b^2)t^2,$$ $$(b^2-1)t^2((t+bp)^2+b^2) = (a^2-b^2)\Bigl(t^2+p(2bt-a^2+b^2)((t+bp)^2+b^2)\Bigr),$$ $$(b^2-1)t^4+2bp(2b^2-a^2-1)t^3 +\Bigl(b^2(b^2-1)(p^2+1)+p(a^2-b^2)^2-a^2+b^2\Bigr)t^2$$ $$+2bp(a^2-b^2)(p(a^2-b^2)-b^2(p^2+1))t +pb^2(a^2-b^2)^2)=0,\tag5$$
If formula $(5)$ is correct for the given parameters' values, then it should be reduced to the form of $$g(t)= t^4+2pt^3\pm q^2t^2+2rs+\pm s^2=0.\tag6$$
And the presentation $$g(t) = (t^2+pt)^2 - (rt+s)^2,$$ taking in account $$q^2 = q^2(\cos^2\beta+\sin^2\beta)= q^2(\cosh^2\gamma-\sinh^2\gamma),$$ allows to obtain unknown $\gamma$ via suitable cubic equation and to deal with the quadratic ones.
Proposed approach looks hard, but I do not see more effective way.