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copied!<p>In a 3D sense you are first concerned with not with a circle but with the plane where the circle lies on. Then you can find the point of intersection between the ray (line) and the plane (disk).</p> <p>I like to use homogeneous coordinates for point, planes and lines and I hope you are familiar with vector dot <code>·</code> and cross products <code>×</code>. Here is the method:</p> <p>Plane (disk) is defined by a point vector <code>r=[rx,ry,rz]</code> and a normal direction vector <code>n=[nx,ny,nz]</code>. Together they form a plane <code>W=[W1,W2]=[n,-r·n]</code>.</p> <p>Line (ray) is defined by two point vectors <code>r_A=[rAx,rAy,rAz]</code> and <code>r_B=[rBx,rBy,rBz]</code>. Together they form the line <code>L=[L1,L2]=[r_B-r_A, r_A×r_B]</code></p> <p>The intersecting Point is defined by <code>P=[P1,P2]=[L1×W1-W2*L2, -L2·W1]</code>, or expanded out as</p> <pre><code>P=[ (r_B-r_A)×n-(r·n)*(r_A×r_B), -(r_A×r_B)·n ] </code></pre> <p>The coordinates for the point are found by <code>r_P = P1/P2</code> where <code>P1</code> has three elements and <code>P2</code> is scalar.</p> <p>Once you have the coordinates you check the distance with the center of the circle by <code>d=sqrt((r_p-r)·(r_p-r))</code> and checking <code>d<=R</code> where <code>R</code> is the radius of the circle. Note the difference in notation between a scalar multiplication <code>*</code> and a dot product <code>·</code></p> <p>If you know for sure that the circles lie on the ground (<code>r=[0,0,0]</code>) and face up (<code>n=[0,0,1]</code>) then you can make a lot of simplifications to the above general case.</p> <p>[ref: <a href="http://www.flipcode.com/archives/Introduction_To_Plcker_Coordinates.shtml" rel="nofollow">Plucker Coordinates</a>]</p> <p><strong>Update:</strong></p> <p>When using the ground (with +Z up) as the plane (where circles lie), then use <code>r=[rx,ry,0]</code> and <code>n=[0,0,1]</code> and the above intersection point simplifies to</p> <pre><code>r_p = [ rBy-rAy, rAx-rBx, 0] / (rAy*rBx-rAx*rBy) </code></pre> <p>of which you can check the distance to the circle center.</p>

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