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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
  
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Smile Re: ortho-implicit circle - 02-25-2008, 03:15 PM

This reminds me of the number 10 written as a circle with the one vertical at its centre.


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