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Mrs. Miniver's problem is a geometry problem about the area of circles. It asks how to place two circles and of given radii in such a way that the lens formed by intersecting their two interiors has equal area to the symmetric difference of and (the area contained in one but not both circles). [1]
In plane geometry, a lune (from Latin luna 'moon') is the concave-convex region bounded by two circular arcs. [1] It has one boundary portion for which the connecting segment of any two nearby points moves outside the region and another boundary portion for which the connecting segment of any two nearby points lies entirely inside the region.
The value of the two products in the chord theorem depends only on the distance of the intersection point S from the circle's center and is called the absolute value of the power of S; more precisely, it can be stated that: | | | | = | | | | = where r is the radius of the circle, and d is the distance between the center of the circle and the ...
The center lens of the 2-circle figure is called a vesica piscis, from Euclid. Two circles are also called Villarceau circles as a plane intersection of a torus. The areas inside one circle and outside the other circle is called a lune. The 3-circle figure resembles a depiction of Borromean rings and is used in 3-set theory Venn diagrams.
Any initial two disjoint given circles can be rendered concentric as follows. The radical axis of the two given circles is constructed; choosing two arbitrary points P and Q on this radical axis, two circles can be constructed that are centered on P and Q and that intersect the two given circles orthogonally. These two constructed circles ...
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 ...
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 ...
Monge's theorem states that the three such points given by the three pairs of circles always lie in a straight line. In the case of two of the circles being of equal size, the two external tangent lines are parallel. In this case Monge's theorem asserts that the other two intersection points must lie on a line parallel to those two external ...