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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle

Let the circle \(x^{2}+y^{2}=4\) intersect x-axis at the points \(A(a,0), a>0\) 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 such that \((\alpha-\beta)=\frac{\pi}{2}.\) Then the point of intersection of AQ and BP lies on:

  1. A \(x^{2}+y^{2}-4y-4=0\)
  2. B \(x^{2}+y^{2}-4x-4=0\)
  3. C \(x^{2}+y^{2}-4x-4y=0\)
  4. D \(x^{2}+y^{2}-4x-4y-4=0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(x^{2}+y^{2}-4y-4=0\)

Step-by-step Solution

Detailed explanation

Let point of intersection \(R ( h , k )\) \(m_{B R}=m_{B P} \Rightarrow \frac{k}{h+2}=\frac{2 \sin \alpha}{2 \cos \alpha+2} \Rightarrow \frac{k}{h+2}=\tan \frac{\alpha}{2}\)…
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