确定圆上的矢量点 [英] Determining Vector points on a circle

问题描述

我有一个问题，就像是trig和pre calc.我们给定了OA的向量点(6 0)，向量OB(-6 0)和OC(5 root11)，它们都是半径为6的圆上的点.点在圆上?

I have a problem it is like trig and pre calc. We are given vector points for OA which is (6 0) and vector OB (-6 0) and OC (5 root11) and they are all points on a circle with a radius of 6. How exactly do you determine or show that these points lie on the circle?

推荐答案

对于不相互共线的任何三个点，都有一个与集合中每个成员等距的点.如果您正在学习三角学，则在三个任意点(而不是半径)的情况下找到该点，将是一个更具启发性的项目.

For any three points which are not mutually co-linear, there is a point equidistant from each member of the set. Finding this point given three arbitrary points (and not given the radius) is a much more informative project if you're learning trigonometry.

在您的示例中，解决方案很简单:x ² + y ² = r ² .您提供的每个点的值都使得它们的平方和等于36，这与半径6的圆所期望的一样.您的问题特别容易，因为所讨论的圆已经位于原点的中心.

In your example, the solution is simple: x² + y² = r². Each of the points you provided have values such that the sum of their squares equals 36, as expected for a circle of radius 6. Your problem is especially easy because the circle in question is already centered on the origin.

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