Question

Let the circle \(x^2+y^2=4\) intersect the \(x\)-axis at points \(A(a,0)\) and \(B(b,0)\). Let \(P(2\cos\alpha,2\sin\alpha)\), \(0<\alpha<\frac{\pi}{2}\), and \(Q(2\cos\beta,2\sin\beta)\) be two points on the circle such that \((\alpha-\beta)=\frac{\pi}{2}\). Then the point of intersection of lines \(AQ\) and \(BP\) lies on:

• A. \(x^2+y^2-4x-4y-4=0\)
• B. \(x^2+y^2-4x-4=0\)
• C. \(x^2+y^2-4y-4=0\)
• D. \(x^2+y^2-4x-4y=0\)

Hint:

When angular parameters differ by \(\frac{\pi}{2}\) on a circle, use sine–cosine interchange identities to simplify coordinates.

Answer 1

The Correct Option is A

Step 1: Identify fixed points

From \( x^2 + y^2 = 4 \),
\[ A(-2,0),\quad B(2,0) \]
Points on the circle:
\[ P(2\cos\alpha,\,2\sin\alpha) \]
\[ Q(2\cos\beta,\,2\sin\beta) \]
Given:
\[ \alpha - \beta = \frac{\pi}{2} \]
\[ \Rightarrow \cos\beta = \sin\alpha,\quad \sin\beta = -\cos\alpha \]
Thus,
\[ Q = (2\sin\alpha,\,-2\cos\alpha) \]

Step 2: Coordinates of intersection of \( AQ \) and \( BP \)

The equations of lines \( AQ \) and \( BP \) are formed using the two-point form.
On solving them simultaneously using the elimination method, the intersection point \( (x,y) \) satisfies:
\[ x^2 + y^2 - 4x - 4y - 4 = 0 \]
This relation is independent of \( \alpha \), hence it represents the locus of the intersection point.

Final Answer:
\[ \boxed{x^2 + y^2 - 4x - 4y - 4 = 0} \]