Question

A circle \(x^2+y^2=4\) intersects the \(x\)-axis at \(A(-2,0)\) and \(B(2,0)\). If two variable points \(P(2\cos\alpha,\,2\sin\alpha)\) and \(Q(2\cos\beta,\,2\sin\beta)\) vary on the circle such that \(\alpha-\beta=\dfrac{\pi}{2}\), then find the locus of the point of intersection of lines \(AP\) and \(BQ\).

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

Hint:

For locus problems with moving points on a circle, use trigonometric parametrization and eliminate the parameter after intersecting the relevant lines.

Answer 1

The Correct Option is A

Step 1: Coordinates of points
\[ A(-2,0),\quad B(2,0) \] \[ P(2\cos\alpha,\,2\sin\alpha),\quad Q(2\cos\beta,\,2\sin\beta),\quad \alpha-\beta=\frac{\pi}{2} \] Thus, \[ \cos\beta=\sin\alpha,\quad \sin\beta=-\cos\alpha. \]
Step 2: Equations of lines
Line \(AP\): \[ \frac{y-0}{x+2}=\frac{2\sin\alpha}{2\cos\alpha+2} \;\Rightarrow\; y=\frac{\sin\alpha}{1+\cos\alpha}(x+2) \] Line \(BQ\): \[ \frac{y-0}{x-2}=\frac{2\sin\beta}{2\cos\beta-2} =\frac{-2\cos\alpha}{2\sin\alpha-2} \;\Rightarrow\; y=\frac{-\cos\alpha}{\sin\alpha-1}(x-2) \]
Step 3: Intersection point
Let the intersection be \((x,y)\). Equating the two expressions for \(y\) and simplifying eliminates \(\alpha\), yielding: \[ x^2+y^2-4y-4=0 \]
Final Answer:
\[ \boxed{x^2+y^2-4y-4=0} \]