Intersect the circle with the two lines and the circle forming the sector's boundary; in each case, if there are intersections, check whether they lie within the part (line segment or arc) forming the boundary. If you don't find intersections, the sector and circle are either disjoint, or one is contained in the other; you can check for the latter case by determining whether the centre of either circle is inside the other object.
[Edit in response to the comment:]
The sector is bounded by two line segments and a circular arc, which are parts of two lines and a circle, respectively. To intersect a line given by $\vec n\cdot\vec x=c$ (with $\vec n$ a unit normal vector) and a circle given by $(\vec x-\vec x_0)^2=r^2$, find the (signed) distance $d=\vec n\cdot\vec x_0-c$ of the circle's origin from the line. If $d\gt r$ there are nor intersections; if $d\le r$, the intersections are at
$$\vec x_0-d\vec n\pm\sqrt{r^2-d^2}\,\vec y\;,$$
where $y$ is a unit vector with $\vec n\cdot\vec y=0$, i.e. in either direction along the line.
To intersect two circles given by $(\vec x-\vec x_1)^2=r_1^2$ and $(\vec x-\vec x_2)^2=r_2^2$, subtract the two equations to obtain
$$2(\vec x_2-\vec x_1)\cdot\vec x=r_1^2-r_2^2-\vec x_1^2+\vec x_2^2\;,$$
which is an equation for a line perpendicular to the line connecting the two centres, which you can intersect with one of the circles as described above to find the intersections of the circles.
The expected distance from a unit circle to a randomly chosen point inside the circle is given by
$$
\frac{1}{\pi}\int_{0}^{1}(1-r)(2\pi r)dr=\left(r^2-\frac{2}{3}r^3\right)\bigg\vert_{0}^{1}=\frac{1}{3};
$$
as a function of the area of the circle, then, the expected distance is
$$
{d}_{\text{circle}}(A)=\frac{1}{3\sqrt{\pi}}\sqrt{A} \approx 0.1881 \sqrt{A}
$$
(which holds for circles of any size). For a square of side length $2$, the expected distance from the square to a random interior point (which can be calculated by considering a single quadrant) is
$$
\int_{0}^{1}y(2-2y)dy = \left(y^2-\frac{2}{3}y^3\right)\bigg\vert_{0}^{1}=\frac{1}{3}
$$
as well, so
$$
d_{\text{square}}(A)=\frac{1}{6}\sqrt{A} \approx 0.1667 \sqrt{A}
$$
for an arbitrary square. Finally, for a $2\times 4$ rectangle, you need to consider the short and long sides differently. Essentially you have two $2 \times 1$ end caps that behave like the square (so the average distance is $1/3$), and a $2\times 2$ central block for which the average distance is just $1/2$. The two components have equal areas, so the overall average distance is the average of $1/2$ and $1/3$, or $5/12$. Since the area of the entire rectangle is $8$, we have
$$
d_{\text{rect}}(A)=\frac{5}{24\sqrt{2}}\sqrt{A} \approx 0.1473 \sqrt{A}
$$
for any rectangle with aspect ratio $2$.
If it helps, you can think of "unrolling" each shape, while preserving its area, so that the set of points at distance $d$ from the perimeter lies along $y=d$. For the circle and the square (and for any regular polygon), this gives a triangle with base equal to the original shape's perimeter. For the rectangle, though, it gives a trapezoid, because the points maximally distant from the perimeter are a line segment, not a single point.
Best Answer
Let's think first why you need to rotate at that particular distance. It is done in such a way that when you take the cross section, in the radial direction, you get the desired figure. You also have the constraint that your cross section is touching the big cylinder.
It does not matter what your shape is. You inscribe it in a rectangle, with the height equal to the distance between the largest and the closest point to the cylinder. Rotate the rectangle as before, and your desired figure will be rotated accordingly.