Let $B$ and $C$ be two fixed points of a given circle and $A$ a variable point lying on major arc of

Question

Let $B$ and $C$ be two fixed points of a given circle and $A$ a variable point lying on major arc of this circle as shown in figure. The locus of feet of perpendiculars dropped from the midpoint of $AB$ on $AC$ is:

Accepted Answer

[Key Concept: Apply the relevant inscribed-angle / circle condition to this option set.]Draw BD ⟂ AC, meeting AC at D.SinceM is the midpoint of AB, and MP ⟂ AC while BD ⟂ AC,we have MP ∥ BD.Therefore, in triangle ABD, by the midpoint theorem:P is the midpoint of AD.Hence, PD = AD/2Now, in right triangle ABD, tan θ = BD/ADConsider right triangle BPD: tan ∠BPD = BD/PD = BD/(AD/2)= 2BD/AD = 2 tan θSince P, D and C are collinear, ∠BPC = tan⁻¹(2 tan θ)Thus, for every position of A, the point P sees the fixed chord BC under the constant angle tan⁻¹(2 tan θ).Therefore, the locus of P is a circle through B and C such that BC subtends the angle tan⁻¹(2 tan θ) at its circumference.

Question

Options

Solution

More Circle Questions