Average distance planar circle to point 3d maybe

[Guest]

There is a 3d space, x, y and z, a point Q and a circle on a plane with center P and any number of outer points may be defined and area and radius are known and Ax+By+Cz=D is all known and a local 2d coordinates system on plane is defined, x0, y0, and we can convert between 3d and 2d coordinates for anything on plane but Q may not be on plane, how to determine average distance of Q from circle with center P's points, all of them? Does say P1 is x on plane way from P and P2 like y with that help? Can you connect with me, maybejosiah@aol.com if you answer?

[shyamjayakannan]

Screenshot 2026-03-14 230109.png (71.82 KiB) Viewed 90 times

Consider the image shown above. We will first look at the situation when all points lie on a plane. So, Q also lies in the plane of the circle with center P. To calculate the average distance, we only look at the upper semi - circle because points on the lower one will have identical distances from Q. Now, we divide the upper semi - circle into [tex]n[/tex] equidistant points and [tex]\theta=\frac{\pi}{n}[/tex] as shown in the figure. Let us look at the distance from the first point:

[tex]QD=\sqrt{r^2+d^2-2rd\cos\theta}=\sqrt{r^2+d^2-2rd\cos\frac{\pi}{n}}[/tex]. So, the average distance for the upper semicircle = [tex]\frac{1}{n}\sum_{k=1}^n\sqrt{r^2+d^2-2rd\cos\frac{k\pi}{n}}[/tex].

Since we need to find this for all points, it becomes = [tex]\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n\sqrt{r^2+d^2-2rd\cos\frac{k\pi}{n}}=\int\limits_0^1\sqrt{r^2+d^2-2rd\cos{\pi x}}\ dx[/tex].

Shifting this to 3D, the value changes to [tex]\int\limits_0^1\sqrt{a^2+r^2+d^2-2rd\cos{\pi x}}\ dx[/tex], where [tex]a[/tex] is the distance between Q and the plane.