If you're not willing to use other libraries and only tkinter, I don't think there's a built in function in tkinter that will allow it. I am also not sure why everything must be done in tkinter as it's not at all unusual for programs to use multiple libraries. Personally, given tikinter's limitations, I would use pygame to track polygons and their intersection but would never draw them. Short of using a third library (tikinter, python default, and other), there is one other approach. I mean it's really the only appraoch.
good ole fashioned math.
https://algs4.cs.princeton.edu/93intersection/
Here's some documentation on how to go about doing that. I wish you luck.
EDIT:
I think I accidentally found the answer to your question using math while studying for some other stuff. Still no way IN tkinter but here's a math explanation.
circle with center (x,y) and radius r
Polygon with z number of sides with x*2-1 number of points(x,y)
If you take iterate the lines of the polygon and put them through the following maths
Line J (x1,y1)(x2,y2)
m of Line J = (y1-y2)/(x1-x2)
create line P from circle center to P1 of LineJ
create line O from circle center to P2 of LineJ
Now we have a triangle
take the inverse cosine and length of P and O to get angle of the triangle you just made.
Make a right triangle by bisecting the triangle with line K starting at circle center and going out at the angle you just calculated.
Now you have line P and 1/2 angle of line P to line K
Now to find the intercept of that mid angle line
Tan(1/2 angle) = slope or m of the new line
using the x,y of the circle center calculate the slope intercept formula y=mx+b and get b
Now take the slope intercept formula for line J and set it equal to the slope intercept of the new line
line J (mx + b) = y = line P (mx + b)
Solve for y
Then plug y in the slope intercept for either and solve for x.
Once you've done this you have 4 points. The three points of the triangle, the point that makes a perpendicular line to center of circle from line J.
If any 3 of those points to the center of the circle in distance is smaller than r, they overlap. If they are are all > r then they don't overlap.