View Single Post
ortho-implicit circle
Old
  (#1 (permalink))
AntonioLao
Raider of the lost time
AntonioLao is a name known to allAntonioLao is a name known to allAntonioLao is a name known to all
 
AntonioLao's Avatar
 
Status: Offline
Posts: 5,613
Thanks Given: 790
Thanked 180x in 174 Posts
Join Date: Nov 2003
Rep Power: 80
   
ortho-implicit circle - 02-25-2008, 03:06 PM

In a Cartesian system, the nonfunctional equation of a circle is given as x+y=R. This implies that the center of the circle is at the origin (0, 0) with radius R. By implicit differentiation, the derivative is found as negative ratio of x over y: dy/dx=-x/y. Since dy/dx is the slope of the tangent line at point (x, y), it is equivalent to the negative reciprocal of the slope of the radius terminating at point (x, y) which is simply y/x, proving that any tangent line to a circle at a point is always perpendicular to the radius terminating at that point. For centers located at arbitrary points (h, k), the equation of the circle becomes (x-h)+(y-k)=R with derivative dy/dx=-(x-h)/(y-k).


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
Reply With Quote