Exploring Analyic Geometry with Mathematica®

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Construction of Two Related Circles

tnlncir.html

Exploration

Prove that if OP and OQ are the tangent lines from (0,0) to the circle

"tnlncir_1.gif"

then the equation of the circle OPQ is

"tnlncir_2.gif".

Approach

Create the circle from the given quadratic and construct the polar (line) of the origin with respect to the circle. Intersect the polar with the circle to find P and Q. Construct a circle through O, P and Q and find its 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 origin point and the circle from the given equation.

Clear[g,f,c];
P0=Point2D[0,0];
C1=Circle2D[Quadratic2D[1,0,1,2g,2f,c]] //Simplify

"tnlncir_3.gif"

Construct the polar line.

L1=Line2D[P0,C1] //Simplify

"tnlncir_4.gif"

Find the intersection points.

pts=Points2D[L1,C1] //FullSimplify

"tnlncir_5.gif"

Construct the circle through the three points.

C2=Circle2D[P0,pts[[1]],pts[[2]]] //FullSimplify

"tnlncir_6.gif"

Convert the circle to an equation.

Clear[x,y];
Equation2D[Quadratic2D[C2]//Simplify,{x,y}]

"tnlncir_7.gif"

Discussion

Construct the circle related to "tnlncir_8.gif".

P0=Point2D[0,0];
C1=Circle2D[Quadratic2D[x^2+y^2-6x-4y+12==0,{x,y}]];
L1=TangentLines2D[P0,C1];
P1=Point2D[First[L1],C1];
P2=Point2D[Last[L1],C1];
C2=Circle2D[Quadratic2D[x^2+y^2-3x-2*y==0,{x,y}]];
Sketch2D[{P0,C1,L1,P1,P2,C2}]

"tnlncir_9.gif"

Graphics saved as "tnlnci01.eps".


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