# 3D Circle/ground intersection

This one stumps me: A circle in 3D space given by its center = $(0.15, 0.5, 1.0)$, its radius $=64$ and an orientation vector that points away from the circle's plane $(0.251, -0.796, 0.551)$

How would I go about finding the $2$ point coordinates where it touches the ground plane?

As a I am math beginner (and always will be), I have given some example numbers because I hoped I'll understand better when I see how they're transformed into the solution.

I'll reply to myself here in case I find the solution.

Is the ground plane the plane $z=0$?

The plane of the circle is given by $r. \left( \begin{array} \\ 0.251\\ -0.796\\ 0.551 \end{array} \right) = \left( \begin{array} \\ 0.15\\ 0.5\\ 1.0 \end{array} \right) . \left( \begin{array} \\ 0.251\\ -0.796\\ 0.551 \end{array} \right) =0.19065$

The problem is asking for a point $(x,y,0)$ that is on the plane i.e.

$\left( \begin{array} \\ x\\ y\\ 0 \end{array} \right).\left( \begin{array} \\ 0.251\\ -0.796\\ 0.551 \end{array} \right)=0.19065$

and of distance $64$ from the centre, i.e.

$\left| \begin{array} \\ x-0.15\\ y-0.5\\ -1.0 \end{array} \right|=64$

$x^2-0.3x+0.0225+y^2-y+0.25+1=4096$

$x=\frac{0.19065+0.796y}{0.251}=0.759562+3.171315y$

$10.057237y^2+4.817621y+0.576934-0.227868-0.951395+0.00225+y^2-y+1.25=4096$

$11.057237y^2+3.817621y-4095.3501=0$

$y=\frac{-3.817621 \pm \sqrt{3.817621^2+4\times 11.057237 \times 4095.3501}}{2 \times 11.057237}=-19.4186,19.0733$

The points are $(-60.82,-19.42,0)$ and $(61.25,19.07,0)$

• Thank you! This is the first internet page I've printed out in years :) -Dizzy May 23 '13 at 0:59