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A black circle with two white circles diagonally adjacent on the same side must have the loop heading away from that side. If not, and it went between the white circles instead, then the white circles would be parallel to that section of the loop, and make it impossible to complete the black circle.
In the special cases of one of the diagonals or sides being a diameter of the circle, this theorem gives rise directly to the angle sum and difference trigonometric identities. [17] The relationship follows most easily when the circle is constructed to have a diameter of length one, as shown here.
Let O 1 and O 2 be the centers of the two circles, C 1 and C 2 and let r 1 and r 2 be their radii, with r 1 > r 2; in other words, circle C 1 is defined as the larger of the two circles. Two different methods may be used to construct the external and internal tangent lines. External tangents Construction of the outer tangent
The cosine rule may be used to give the angles A, B, and C but, to avoid ambiguities, the half angle formulae are preferred. Case 2: two sides and an included angle given (SAS). The cosine rule gives a and then we are back to Case 1. Case 3: two sides and an opposite angle given (SSA). The sine rule gives C and then we have Case 7. There are ...
Given a unit sphere, a "spherical triangle" on the surface of the sphere is defined by the great circles connecting three points u, v, and w on the sphere (shown at right). If the lengths of these three sides are a (from u to v ), b (from u to w ), and c (from v to w ), and the angle of the corner opposite c is C , then the (first) spherical ...
In geometry, the limiting points of two disjoint circles A and B in the Euclidean plane are points p that may be defined by any of the following equivalent properties: The pencil of circles defined by A and B contains a degenerate (radius zero) circle centered at p. [1] Every circle or line that is perpendicular to both A and B passes through p ...
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In general, the same inversion transforms the given line L and given circle C into two new circles, c 1 and c 2. Thus, the problem becomes that of finding a solution line tangent to the two inverted circles, which was solved above. There are four such lines, and re-inversion transforms them into the four solution circles of the Apollonius problem.