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 data of your problem is: $P=(a,b)$ outside point; $C=(c,d)$ center of circle; $r$ radius of circle. Let Q=(x,y) be your generic point and $D$ the searched distance. You have two equations to use:
$$(1)....( PQ)^2=(x-a)^2+(y-b)^2= D^2$$
$$(2)....(x-c)^2+(y-d)^2=r^2$$
Hence (1)-(2) gives $$D^2=2(c-a)x+2(d-b)y+a^2+b^2-c^2-d^2+r^2$$
Where $D$ is function only of $x$ and $y$ as must be answer.
Best Answer
We have that indicating with:
$$\alpha = 2\arccos \left(\frac R d\right)$$