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Reciprocal of a Circle

reccir.html

Exploration

Given a circle "reccir_1.gif" show that its polar reciprocal in the auxiliary conic "reccir_2.gif" is given by the quadratic

"reccir_3.gif".

Furthermore, show that Q is an ellipse if the origin (0,0) is inside C; a parabola, if the origin is on C; and a hyperbola, if the origin is outside C.

Approach

Create a general circle and the auxiliary conic. Construct five points on the circle. Construct five tangent lines at the points. Construct reciprocals of the lines (five points). Construct a quadratic through five points. Examine the discriminant of the quadratic.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Solution

Create a general circle and an auxiliary circle.

Clear[h,k,r];
cir1=Circle2D[{h,k},r];
c1=Circle2D[{0,0},1];

Define five points on the circle.

pts=Map[
  Point2D[cir1[#]]&,
  {0,Pi/4,Pi/2,Pi,3Pi/2}] //Simplify

"reccir_4.gif"

Determine the tangent lines at the points.

lns1=Map[Line2D[#,cir1]&,pts] //Simplify

"reccir_5.gif"

Define the reciprocal function.

Reciprocal2D[
   Line2D[A1_,B1_,C1_],
   Circle2D[{0,0},1]] :=
Point2D[{-A1/C1,-B1/C1}];

Find the reciprocal points.

pts1=Map[Reciprocal2D[#,c1]&,lns1] //Simplify

"reccir_6.gif"

Find the quadratic through the points.

q1=Quadratic2D[Sequence @@ pts1] //Simplify;
Map[(-1*#/2)&,q1]

"reccir_7.gif"

Discussion

Examine the discriminant, d.

disc1=q1[[2]]^2-4*q1[[1]]*q1[[3]] //Simplify

"reccir_8.gif"

If d<0 the quadratic is an ellipse and (0,0) is inside the circle; if d=1 the quadratic is a parabola and (0,0) is on the circle; and if d>1 the quadratic is a hyperbola and (0,0) is outside the circle.


Copyright © 1999-2007 Donald L. Vossler, Descarta2D Publishing
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