Let the circle x 2 + y 2 = 4 intersect x-axis at the points A ( a , 0 ) , a 0 a n d B ( b , 0 ) . Let P ( 2 cos , ? 2 sin ) 0 2 a n d Q ( 2 cos , ? 2 sin ) be two points such that = 2 . Then the point of intersection of AQ and BP lies

Question

Let the circle $x^{2} + y^{2} = 4$ intersect x-axis at the points $A (a, 0), a > 0 and B (b, 0) .$ Let $P (2 \cos α, 2 \sin α) 0 < α < \frac{π}{2} and Q (2 \cos β, 2 \sin β)$ be two points such that $(α - β) = \frac{π}{2}$ . Then the point of intersection of AQ and BP lies on : [2026]

Smart Achievers · Accepted Answer

(2)

$Let point of intersection R (h, k)$

$m_{B R} = m_{B P} \Rightarrow \frac{k}{h + 2} = \frac{2 \sin α}{2 \cos α + 2} \Rightarrow \frac{k}{h + 2} = \tan \frac{α}{2}$

$m_{A R} = m_{A Q} \Rightarrow \frac{k}{h - 2} = \frac{2 \sin β}{2 \cos β - 2} = \frac{\sin β}{\cos β - 1} = - \cot \frac{β}{2}$

$\frac{α}{2} - \frac{β}{2} = \frac{π}{4}$

$\tan (\frac{α}{2} - \frac{β}{2}) = \tan \frac{π}{4} = 1$

$\frac{\tan \frac{α}{2} - \tan \frac{β}{2}}{1 + \tan \frac{α}{2} \tan \frac{β}{2}} = 1$

$\frac{\frac{k}{h + 2} + \frac{h - 2}{k}}{1 + (\frac{k}{h + 2}) (\frac{2 - h}{k})} = 1$

$\Rightarrow \frac{k^{2} + h^{2} - 4}{\frac{k (h + 2)}{\frac{4}{h + 2}}} = 1$

$\Rightarrow \frac{h^{2} + k^{2} - 4}{4 k} = 1$

$\Rightarrow x^{2} + y^{2} - 4 y - 4 = 0$

Solution

Q.
Let the circle $x^{2} + y^{2} = 4$ intersect x-axis at the points $A (a, 0), a > 0 and B (b, 0) .$ Let $P (2 \cos α, 2 \sin α) 0 < α < \frac{π}{2} and Q (2 \cos β, 2 \sin β)$ be two points such that $(α - β) = \frac{π}{2}$ . Then the point of intersection of AQ and BP lies on : [2026]

1 $x^{2} + y^{2} - 4 x - 4 = 0$

2 $x^{2} + y^{2} - 4 y - 4 = 0$

3 $x^{2} + y^{2} - 4 x - 4 y - 4 = 0$

4 $x^{2} + y^{2} - 4 x - 4 y = 0$

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Solution

Q. Let the circle x2+y2=4 intersect x-axis at the points A(a,0), a>0 and B(b,0). Let P(2cosα, 2sinα) 0<α<π2 and Q(2cosβ, 2sinβ) be two points such that (α-β)=π2. Then the point of intersection of AQ and BP lies on : [2026]

1 x2+y2−4x−4=0

2 x2+y2−4y−4=0

3 x2+y2−4x−4y-4=0

4 x2+y2−4x−4y=0