Exploring Analyic Geometry with Mathematica®

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Circular Conic Arc

cacircle.html

Exploration

Show that the conic arc with control points (0,0), (a,b) and (2a,0) will be a circular arc if

"cacircle_1.gif".

Approach

Create the conic arc and find the quadratic associated with it. Force the quadratic's coefficients to represent a circle and solve for ρ.

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 conic arc.

Clear[a,b,p];
ca1=ConicArc2D[{0,0},{a,b},{2a,0},p];

Construct the quadratic associated with the conic arc.

Q1=Quadratic2D[ca1];

Extract and simplify the coefficients.

{a1,b1,c1,d1,e1,f1}=Map[Together,List @@ Q1]

"cacircle_2.gif"

Find ρ that makes the quadratic a circle.

ans1=FullSimplify[Solve[a1==c1,p],Assumptions->{a>0}]

"cacircle_3.gif"

ans1=Solve[a1==c1,p] //Simplify

"cacircle_4.gif"

Use the positive result.

ans2=FullSimplify[Last[ans1][[1]],Assumptions->{a>0}]

"cacircle_5.gif"

Confirm the equality of the two solution forms.

FullSimplify[(p /. ans2) - a*(-a+Sqrt[a^2+b^2])/b^2]

"cacircle_6.gif"

Discussion

A numerical example with a=6 and b=40.

ca2=ca1 //. {a->6,b->20,ans2};
Sketch2D[{ca2}]

"cacircle_7.gif"

Graphics saved as "cacirc01.eps".


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