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The theorem may also be proven using trigonometry: Let O = (0, 0), A = (−1, 0), and C = (1, 0). Then B is a point on the unit circle (cos θ, sin θ). We will show that ABC forms a right angle by proving that AB and BC are perpendicular — that is, the product of their slopes is equal to −1. We calculate the slopes for AB and BC:
In Euclidean plane geometry, a tangent line to a circle is a line that touches the circle at exactly one point, never entering the circle's interior. Tangent lines to circles form the subject of several theorems , and play an important role in many geometrical constructions and proofs .
Gauss's circle problem asks how many points there are inside this circle of the form (,) where and are both integers. Since the equation of this circle is given in Cartesian coordinates by x 2 + y 2 = r 2 {\displaystyle x^{2}+y^{2}=r^{2}} , the question is equivalently asking how many pairs of integers m and n there are such that
Ptolemy's Theorem yields as a corollary a pretty theorem [2] regarding an equilateral triangle inscribed in a circle. Given An equilateral triangle inscribed on a circle, and a point on the circle. The distance from the point to the most distant vertex of the triangle is the sum of the distances from the point to the two nearer vertices.
Secant-, chord-theorem. For the intersecting secants theorem and chord theorem the power of a point plays the role of an invariant: . Intersecting secants theorem: For a point outside a circle and the intersection points , of a secant line with the following statement is true: | | | | = (), hence the product is independent of line .
In geometry, Descartes' theorem states that for every four kissing, or mutually tangent circles, the radii of the circles satisfy a certain quadratic equation. By solving this equation, one can construct a fourth circle tangent to three given, mutually tangent circles. The theorem is named after René Descartes, who stated it in 1643.
Consider the great circle that contains the side BC. This great circle is defined by the intersection of a diametral plane with the surface. Draw the normal to that plane at the centre: it intersects the surface at two points and the point that is on the same side of the plane as A is (conventionally) termed the pole of A and it is denoted by A'.
The figure at the right shows a sector of a circle with radius 1. The sector is θ/(2 π) of the whole circle, so its area is θ/2. We assume here that θ < π /2. = = = = The area of triangle OAD is AB/2, or sin(θ)/2.