Inversion

inverse.html

Exploration

A point P'(x',y') is said to be the inverse of a point P(x,y) in the circle

if points O(h,k), P and P' are collinear and . Using this definition show that

A.  The coordinates of P'(x',y') are

and .

B.  If the circle of inversion is , the coordinates of P' are

and .

C.  If the circle of inversion is , the inverse of the line , assuming L does not pass through the origin, is the circle

.

D.  If the circle of inversion is , the inverse of the line , assuming L passes through the origin (), is L itself.

E.  If the circle of inversion is , the inverse of the circle is

where .

F. If the circle of inversion is , the inversion of , which passes through the origin, is the line . L is parallel to the tangent line to C through the origin. The equation of the tangent line is .

G.  Inversion is clearly a non-rigid transformation.

Approach

See the commentary below.

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

Use the definition of inversion to find the coordinates of a point (x,y) inverted in the circle .  This is the solution to proposition A as stated above.

Clear[x,y,h,k,r];
Coordinates2D[
    Point2D[
       Point2D[h,k],
       Point2D[x,y],
       r^2/Sqrt[(x-h)^2+(y-k)^2]]]

Define a function for inverting coordinates.

Inverse2D[{x_,y_},Circle2D[{h_,k_},r_]] :=
   {h + (r^2*(x-h))/((x-h)^2+(y-k)^2),
    k + (r^2*(y-k))/((x-h)^2+(y-k)^2)}

Here's the inversion in a unit circle at the origin.

invPts=Inverse2D[{x,y},Circle2D[{0,0},1]]

Determine the inverse inversion equations.  This is the solution to proposition B as stated above.

Clear[x1,y1];
eqn1=Solve[{x1,y1}==invPts,{x,y}]

Find the inversion of a line.

Clear[A1,B1,C1];
eq1=A1*x+B1*y+C1 /. First[eqn1]

Clear the denominators of the equations.

eq2=eq1*(x1^2+y1^2) //Simplify

Determine the quadratic (a circle).  This is the solution to proposition C as stated above.

Circle2D[Quadratic2D[eq2,{x1,y1}]]

Find a line passing through the center of inversion (0,0).

eq3=A1*x+B1*y /. First[eqn1]

Clear the denominator.

eq4=eq3 /. {x1^2+y1^2->1}

The line inverts into itself.  This is the solution to proposition D as stated above.

Line2D[eq4,{x1,y1}]

Inversion of a circle.

Clear[h1,k1,r1];
eq5=(x-h1)^2+(y-k1)^2-r1^2 /. First[eqn1]

Clear the denominators.

eq6=eq5*(x1^2+y1^2)^2 //Together;
eq7=eq6[[3]]

Find the circle.  If the resulting denominators are zero, then the circle passes through the center of inversion and the inversion is invalid.  This is the solution to proposition E as stated above.

Circle2D[Quadratic2D[eq7,{x1,y1}]] //Simplify

A circle not passing through the origin.

eq8=(x-h1)^2+(y-k1)^2-(h1^2+k1^2) /. First[eqn1] //Simplify

Clear the denominator and find the line.  This is the solution to the first part of proposition F as stated above.

eq9=Numerator[eq8];
Line2D[eq9,{x1,y1}]

The line is parallel to the tangent at the origin.  This is the solution to the second part of proposition F as stated above.

Line2D[Point2D[0,0],
       Circle2D[{h1,k1},Sqrt[h1^2+k1^2]]]