Hidden circle

Hidden circle — a circle the problem never draws, but a moving point silently traces. Spot it and an "angle constraint" turns into a "circle constraint" — the solution path snaps into focus.

Core recognition

A moving point $P$ + a fixed chord $AB$ (two fixed points) + $\angle APB$ has a fixed value $\Rightarrow$ $P$ lies on a fixed circle.

Inscribed angle $\angle APB$	Locus circle
$= 90°$	Circle with $AB$ as diameter (immediate from Thales converse)
$= \alpha$ (any fixed angle)	Circle with $AB$ as chord and radius $R = \dfrac{AB}{2\sin\alpha}$ (inscribed-angle theorem, Central angle is twice the inscribed angle on the same arc)
$\alpha$ acute / obtuse	$P$ stays on one side of $AB$; acute → major-arc side, obtuse → minor-arc side

The two standard triggers

Trigger 1 · Thales ($\angle = 90°$)

The problem says: "the moving point $P$ is the right-angle vertex of some right triangle, and the two endpoints of the right-angle legs are fixed" → $P$ lies on the circle with the hypotenuse as diameter.

The proof is Thales converse — by the right-triangle median to the hypotenuse (Right triangle: median to hypotenuse = half hypotenuse), $P$'s distance to the midpoint of the hypotenuse is constant, so $P$ lies on the circle centred at that midpoint with radius half the hypotenuse.

Trigger 2 · Inscribed angle ($\angle = \alpha$ fixed)

The problem says: "the moving point $P$ sees a fixed chord $AB$ at a fixed angle $\alpha$" → $P$ lies on a circle with $AB$ as chord and radius $R = AB/(2\sin\alpha)$.

The proof is the converse of Inscribed angles on same arc are equal; angle in a semicircle is right — inscribed angles on the same arc are equal; conversely, if angles subtended by a common chord are equal then the vertices lie on the same arc.

Special-angle table:

$\alpha$	$R$
$30°$	$R = AB$
$45°$	$R = AB / \sqrt 2$
$60°$	$R = AB / \sqrt 3$
$90°$	$R = AB / 2$ (degenerates to Thales)
$120°$	$R = AB / \sqrt 3$
$135°$	$R = AB / \sqrt 2$
$150°$	$R = AB$

Usage (recognise → draw circle → read off the extremum)

A hidden circle usually shows up in a two-step chain:

1. Step 1: identify "angle constraint" as "the locus is a circle" — draw the hidden circle $\Gamma$, mark its centre $O$ and radius $R$;

2. Step 2: read "distance from $P$ to a third point $X$ (or line $\ell$)" off the circle $\Gamma$:

   • $X$ a fixed point: $|PX|_{\max} = |OX| + R$, $|PX|_{\min} = \big||OX| - R\big|$ (equality when $X, O, P$ are collinear)
   • $\ell$ a fixed line: $d(P, \ell)_{\max} = d(O, \ell) + R$, $d(P, \ell)_{\min} = \big|d(O, \ell) - R\big|$ (equality where the perpendicular from $O$ meets the circle)

When to use / mnemonic

• "Moving point + fixed chord + fixed-value subtended angle" ⇒ draw a hidden circle immediately
• "Moving point is the right-angle vertex of some right triangle with fixed leg endpoints" ⇒ hidden circle (Thales variant)
• Compound recognition: the problem disguises the fixed angle as "two fixed leg lengths + a fixed third side + law of cosines gives the included angle" → compute the angle, then apply the same template
• Reverse use: given that a moving point lies on a known circle, prove some angle is constant → equal inscribed angles on the same arc hand it to you (see Inscribed angles on same arc are equal; angle in a semicircle is right)

Pitfalls

• Same-side vs opposite-side branches: the same fixed chord + same fixed angle correspond to two arcs (one on each side of the chord). Which arc $P$ lives on is locked by other constraints in the problem; forgetting the distinction yields a "ghost extremum" from the symmetric solution.
• Range of the subtended angle: $\alpha$ acute → major-arc side; obtuse → minor-arc side; $\alpha = 90°$ degenerates to the whole circle on diameter $AB$ (minus the endpoints $A, B$).
• Spotting that the subtended angle is constant: the problem rarely says so explicitly. Common disguises: fixed side ratios + law of cosines, baked-in figures (square / rectangle vertices give right angles), or the equal base angles of an isosceles triangle after an auxiliary construction.
• Radius formula only holds when the angle is constant: if $\angle APB$ varies with $P$, the locus is not a circle — do not apply the formula.

Applications

To be added.

Related

• Thales: angle on a diameter = 90° / Thales converse — the $\alpha = 90°$ standard trigger
• Central angle is twice the inscribed angle on the same arc / Inscribed angles on same arc are equal; angle in a semicircle is right — the inscribed-angle basis for general $\alpha$ (radius formula + equal angles on the same arc)
• Right triangle: median to hypotenuse = half hypotenuse — equivalent formulation of the Thales variant
• [[miller-theorem]] — extremal use of the hidden circle (maximum viewing angle)
• Melon-and-bean (spiral similarity) — "follower point also lies on a hidden circle" — locus-transfer variant of the hidden circle