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Carlyle Circle

carlyle.html

Exploration

Given a circle, , passing through the three points (0,1), (0,-p) and (s,-p), show that the x-coordinates of the intersection points and of with the x-axis are the roots of the quadratic equation .

Approach

Construct the circle through three points and intersect it with the x-axis. Solve the quadratic equation directly and show that the roots are equal to the x-coordinates of the intersection points.

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

Construct the circle through three points.

Clear[p,s];

C1=Circle2D[

p1=Point2D[{0,1}],

p2=Point2D[{0,-p}],

p3=Point2D[{s,-p}]] //FullSimplify

Intersect the circle with the x-axis.

pts=Points2D[Line2D[0,1,0],C1] //FullSimplify

Solve the quadratic directly which produces the same roots as the x-axis intersections.

Clear[x];

Solve[x^2-s*x-p==0,x]

Discussion

This is a plot of a numerical example with p=2 and s=4.

Sketch2D[{C1,pts,p1,p2,p3} /. {p->2,s->4}]

Graphics saved as "carlyl01.eps".

The intersection points on the x-axis are the same as the roots of the equation.

NSolve[x^2-4*x-2==0,x]

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