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where A is the area of an epicycloid with the smaller circle of radius r and the larger circle of radius kr (), assuming the initial point lies on the larger circle. A = ( − 1 ) k + 3 8 π a 2 {\displaystyle A={\frac {(-1)^{k}+3}{8}}\pi a^{2}}
For example, a circle of radius 2 in a plane, centered on a particular point called the origin, may be described as the set of all points whose coordinates x and y satisfy the equation x 2 + y 2 = 4. Parametric equations
Another proof that uses triangles considers the area enclosed by a circle to be made up of an infinite number of triangles (i.e. the triangles each have an angle of dπ at the centre of the circle), each with an area of β 1 / 2 β · r 2 · dπ (derived from the expression for the area of a triangle: β 1 / 2 β · a · b · sinπ ...
The sum of the squared lengths of any two chords intersecting at right angles at a given point is the same as that of any other two perpendicular chords intersecting at the same point and is given by 8r 2 − 4p 2, where r is the circle radius, and p is the distance from the centre point to the point of intersection. [14]
Kissing circles. Given three mutually tangent circles (black), there are, in general, two possible answers (red) as to what radius a fourth tangent circle can have.In geometry, Descartes' theorem states that for every four kissing, or mutually tangent, circles, the radii of the circles satisfy a certain quadratic equation.
The area-equivalent radius of a 2D object is the radius of a circle with the same area as the object Cross sectional area of a trapezoidal open channel, red highlights the wetted perimeter, where water is in contact with the channel.
A parametric equation for the sphere with radius > and center (,,) can be ... For example, a sphere with diameter 1 m has 52.4% the volume of a cube with edge length ...
Laplace's equation on an annulus (inner radius r = 2 and outer radius R = 4) with Dirichlet boundary conditions u(r=2) = 0 and u(R=4) = 4 sin(5 θ See also: Boundary value problem The Dirichlet problem for Laplace's equation consists of finding a solution φ on some domain D such that φ on the boundary of D is equal to some given function.