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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 ...
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 ...
This easy soup recipe is made with just three ingredients—perfect for a quick and healthy lunch. Plus, this soup has over 20% of the Daily Value of vitamins A and C, two nutrients that are ...
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.
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.
In geometry, a set of Johnson circles comprises three circles of equal radius r sharing one common point of intersection H.In such a configuration the circles usually have a total of four intersections (points where at least two of them meet): the common point H that they all share, and for each of the three pairs of circles one more intersection point (referred here as their 2-wise intersection).
Two circles (C2) centered at B and B', with radius AB, cross again at point C. A circle (C3) centered at C with radius AC meets (C1) at D and D'. Two circles (C4) centered at D and D' with radius AD meet at A, and at O, the sought center of (C). Note: for this to work the radius of circle (C1) must be neither too small nor too large.