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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]
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 ...
The constant chord theorem is a statement in elementary geometry about a property of certain chords in two intersecting circles. The circles k 1 {\displaystyle k_{1}} and k 2 {\displaystyle k_{2}} intersect in the points P {\displaystyle P} and Q {\displaystyle Q} .
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, tangent circles (also known as kissing circles) are circles in a common plane that intersect in a single point. There are two types of tangency : internal and external. Many problems and constructions in geometry are related to tangent circles; such problems often have real-life applications such as trilateration and maximizing the ...
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).
Illustration of Paden–Kahan Subproblem 2. The subproblem yields two solutions in the event that the circles intersect at two points; one solution if the circles are tangential; and no solution if the circles fail to intersect. Let and be two zero-pitch twists with unit magnitude and intersecting axes.
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.