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intersection of two polygons: window test. If one wants to determine the intersection points of two polygons, one can check the intersection of any pair of line segments of the polygons (see above). For polygons with many segments this method is rather time-consuming. In practice one accelerates the intersection algorithm by using window tests ...
Creating the one point or two points in the intersection of a line and a circle (if they intersect) Creating the one point or two points in the intersection of two circles (if they intersect). For example, starting with just two distinct points, we can create a line or either of two circles (in turn, using each point as centre and passing ...
To generate the line that bisects the angle between two given rays [clarification needed] requires a circle of arbitrary radius centered on the intersection point P of the two lines (2). The intersection points of this circle with the two given lines (5) are T1 and T2. Two circles of the same radius, centered on T1 and T2, intersect at points P ...
These two circles determine a pencil, meaning a line L in the P 3 of circles. If the equations of C 0 and C ∞ are f and g, respectively, then the points on L correspond to the circles whose equations are Sf + Tg, where [S : T] is a point of P 1. The points where L meets Z D are precisely the circles in the pencil that are tangent to D.
Step 1 (red): construct a circle with center at P to create points A' and B' on the line AB, which are equidistant from P. Step 2 (green): construct circles centered at A' and B' having equal radius. Let Q and P be the points of intersection of these two circles. Step 3 (blue): connect Q and P to construct the desired perpendicular PQ.
D, the other point of intersection of the two circles, is the reflection of C across the line AB. If C = D (that is, there is a unique point of intersection of the two circles), then C is its own reflection and lies on the line AB (contrary to the assumption), and the two circles are internally tangential.
The distances between the centers of the nearer and farther circles, O 2 and O 1 and the point where the two outer tangents of the two circles intersect (homothetic center), S respectively can be found out using similarity as follows: Here, r can be r 1 or r 2 depending upon the need to find distances from the centers of the nearer or farther ...
The intersection points between any line and the given circle (or given arc of a circle) may be found directly, as can the intersection points between the arcs of two circles, if provided. The Poncelet-Steiner Theorem does not prohibit the normal treatment of circles already drawn in the plane; normal construction rules apply.