Searched circle

To construct a circle centered on a given line (d), through a given point P and intersecting a given line (Δ) in a chord of given length 2a.
Source : Geometriagon

Consider the locus of the circle centers through P and intersecting (Δ) in a constant chord 2a.
Analytically, it is easy to prove that it is a parabola, of which we shall define the geometric properties.
Let O (x, y) the center of such a circle, then M = (x-a, 0) and N = (x+a, 0)
OP² = OM² is written, using P = (0, h), x² + (y - h)² = a² + y², that is :

  2hy = x² + h² - a² 

The vertex S is of course defined as the circumcenter of PMoNo, with Mo,No = (±a, 0).
Analytically (X² = 2pY), the parameter of this parabola is h = PH, and then the focus F is from SF = HP/2 = h/2, and the directrix is at distance PH = -h from focus.

In the applet, point 2a defines the length of the chord.
Point M on Δ defines a current circle.

Sorry, your browsr is not Java compliant

The wanted construction is then the classical construction of the intersection points of a line (d) with a parabola with focus F and directrix δ
In the applet, line (d) is defined from its two blue points.

The general construction fails if P is on line (Δ), but then the construction becomes obvious !
If (d) is perpendicular to (Δ), that is parallel to parabola axis, the applet fails.
There is in this case only one solution, with obvious construction as (d) is the perpendicular bissector of MN.


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