Medial Curve, Point–Circle

mdptcir.html

Exploration

Show that the quadratic equation

where

,

,

,

,

,

and

is equidistant from the point and the circle

.

Approach

Create the point and the circle. Compute the distances to an arbitrary point. Set the distances equal to form the equation.

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 the point and the circle.

Clear[x,y,x1,y1,h2,k2,r2];
P=Point2D[x,y];
p1=Point2D[x1,y1];
c2=Circle2D[{h2,k2},r2];

Compute the distance between the two points.

d1=Distance2D[P,p1]^2

The distance to the circle is either D-r or r-D, where D is the distance from the point to the center of the circle. Squaring removes the ambiguity.

Clear[a,b];
d2=(Distance2D[P,Point2D[c2]]-r2)^2 /.
      (a_+b_)^2->a^2+2a*b+b^2

Simplify the equations .

{ls1=d1-d2[[{1,2,3}]] //Expand,
rs1=d2[[4]]}

Clear[R];
ls2=ls1 /.
   {-h2^2-k2^2+x1^2+y1^2-r2^2->-2*r2^2-R}

Q1=Quadratic2D[ls2^2==rs1^2,{x,y}]

Clear[E1,E2];
a=Factor[Q1[[1]]] //. {
    (E1_-E2_)(E1_+E2_)->E1^2-E2^2,
   (h2-x1)^2->(x1-h2)^2}

b=Factor[Q1[[2]]] /. (h2-x1)(k2-y1)->(x1-h2)(y1-k2)

c=Factor[Q1[[3]]] //. {
(E1_-E2_)(E1_+E2_)->E1^2-E2^2,
(k2-y1)^2->(y1-k2)^2}

d=Factor[Q1[[4]]] //. {
   h2*R-x1*R->R(h2-x1),
   -4(R(h2-x1)-2*r2^2*x1)->
    4(R(x1-h2)+2*r2^2*x1)}

e=Factor[Q1[[5]]] //. {
   k2*R-y1*R->R(k2-y1),
   -4(R(k2-y1)-2*r2^2*y1)->
    4(R(y1-k2)+2*r2^2*y1)}

f=Factor[Q1[[6]]] /.
  4*R*r2^2->4*r2^2*(h2^2+k2^2-(x2^2+y2^2)-r2^2) //Expand;
f1=f[[1]]+Factor[f[[{2,3}]]]

Quadratic2D[a,b,c,d,e,f1]