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Hyperbolic center (Posted on 2024-02-06) Difficulty: 3 of 5
A variable circle in the xy-plane is tangent to the x-axis and meets the y-axis at the points P and Q. If the circle varies in such a way that the length of the segment PQ is always 2, show that the center A of the circle lies on a hyperbola, and find the equation of this hyperbola.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution | Comment 2 of 3 |
For parameter 'a', the equation of the circle can be:
(x-a)^2 + (y - √(1+a^2))^2 = 1+a^2

The center is (a, √(1+a^2))
The radius is √(1+a^2)
The y intercepts are √(1+a^2) ± 1 which puts their separation at 2 units.

The parametric equation for the locus of points that can be the center of the circle is:
x = a
y = √(1+a^2)

Since x = a, substitute
y = √(1+x^2)
y^2 = 1+x^2
y^2 - x^2 = 1 with y>0  which is a hyperbola

See a graphic representation on Desmos:
https://www.desmos.com/calculator/2i1gfg6snv

  Posted by Larry on 2024-02-06 12:36:26
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