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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
Using the equations for lines and circles, one can show that the points at which they intersect lie in a quadratic extension of the smallest field F containing two points on the line, the center of the circle, and the radius of the circle. That is, they are of the form + =, where x, y, and k are in F.
This can be simplified in various ways, to conform to more specific cases, such as the equation = for a circle with a center at the pole and radius a. [ 15 ] When r 0 = a or the origin lies on the circle, the equation becomes r = 2 a cos ( φ − γ ) . {\displaystyle r=2a\cos(\varphi -\gamma ).}
Specifically, stereographic projection from the north pole (0,1) onto the x-axis gives a one-to-one correspondence between the rational number points (x, y) on the unit circle (with y ≠ 1) and the rational points of the x-axis. If ( m / n , 0) is a rational point on the x-axis, then its inverse stereographic projection is the point
A circle circumference and radius are proportional. The area enclosed and the square of its radius are proportional. The constants of proportionality are 2 π and π respectively. The circle that is centred at the origin with radius 1 is called the unit circle. Thought of as a great circle of the unit sphere, it becomes the Riemannian circle.
Equivalently, it is the set of vertices with eccentricity equal to the graph's radius. [3] Thus vertices in the center (central points) minimize the maximal distance from other points in the graph. This is also known as the vertex 1-center problem and can be extended to the vertex k-center problem. Finding the center of a graph is useful in ...
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Because in a continuous function, the function for a sphere is the function for a circle with the radius dependent on z (or whatever the third variable is), it stands to reason that the algorithm for a discrete sphere would also rely on the midpoint circle algorithm. But when looking at a sphere, the integer radius of some adjacent circles is ...