Question

A circle having centre at (0, 0) and radius equal to 'a' meets the x - axis at P and Q. A(α) and B(β) are points on this circle such that α – β = 2γ, where γ is a constant. Then locus of the point of intersection of PA and QB is

(A) x² – y² – 2ay tan γ = a² 

(B) x² + y² – 2ay tan γ = a² 

(C) x² + y² + 2ay tan γ = a² 

(D) x² – y² + 2ay tan γ = a

Answer 1

Correct option (B) x² + y² – 2ay tan γ = a²   

Explanation:

Let the equation of the circle be x² +y² = a²

Similarly, equation of BQ is

Now, we eliminate α β, using (i) and (ii)