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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.
Draw three circumcircles (Miquel's circles) to triangles AB´C´, A´BC´, and A´B´C. Miquel's theorem states that these circles intersect in a single point M, called the Miquel point. In addition, the three angles MA´B, MB´C and MC´A (green in the diagram) are all equal, as are the three supplementary angles MA´C, MB´A and MC´B. [2] [3]
This special line is the radical line of the two circles. Intersection of two circles with centers on the x-axis, their radical line is dark red. Special case = = = : In this case the origin is the center of the first circle and the second center lies on the x-axis (s. diagram).
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 ...
Next to the intersecting chords theorem and the tangent-secant theorem, the intersecting secants theorem represents one of the three basic cases of a more general theorem about two intersecting lines and a circle - the power of point theorem.
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 ...
Apollonius' problem can be extended to construct all the circles that intersect three given circles at a precise angle θ, or at three specified crossing angles θ 1, θ 2 and θ 3; [50] the ordinary Apollonius' problem corresponds to a special case in which the crossing angle is zero for all three given circles.
Let be the intersection of line with and be the intersection of line with . Homothety with center on T A {\displaystyle T_{A}} between Ω A {\displaystyle \Omega _{A}} and Γ {\displaystyle \Gamma } implies that X , Y {\displaystyle X,Y} are the midpoints of Γ {\displaystyle \Gamma } arcs A B {\displaystyle AB} and A C {\displaystyle AC ...