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If = + is the distance from c 1 to c 2 we can normalize by =, =, = to simplify equation (1), resulting in the following system of equations: + =, + =; solve these to get two solutions (k = ±1) for the two external tangent lines: = = + = (+) Geometrically this corresponds to computing the angle formed by the tangent lines and the line of ...
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 ...
Let PQ be a line perpendicular to line OQ defined by angle , drawn from point Q on this line to point P. OQP is a right angle. Let QA be a perpendicular from point A on the x -axis to Q and PB be a perpendicular from point B on the x -axis to P. ∴ {\displaystyle \therefore } OAQ and OBP are right angles.
A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x 3 − 3x + d = 0, where is the value of the cosine function at the one-third angle and d is the known value of the cosine function at the full angle.
The Elements contains the proof of an equivalent statement (Book I, Proposition 27): If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another. As De Morgan [23] pointed out, this is logically equivalent to (Book I, Proposition 16).
This proof utilizes two facts: two lines form a right angle if and only if the dot product of their directional vectors is zero, and; the square of the length of a vector is given by the dot product of the vector with itself. Let there be a right angle ∠ ABC and circle M with AC as a diameter. Let M's center lie on the origin, for easier ...
Suppose S is the common starting point of two rays, and two parallel lines are intersecting those two rays (see figure). Let A, B be the intersections of the first ray with the two parallels, such that B is further away from S than A, and similarly C, D are the intersections of the second ray with the two parallels such that D is further away ...
Poincaré disc: The pink line is ultraparallel to the blue line and the green lines are limiting parallel to the blue line. In hyperbolic geometry, two lines are said to be ultraparallel if they do not intersect and are not limiting parallel. The ultraparallel theorem states that every pair of (distinct) ultraparallel lines has a unique common ...