JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

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 $A Q$ and $B P$ lies on :

A
$x^{2} + y^{2} - 4 y - 4 = 0$
B
$x^{2} + y^{2} - 4 x - 4 = 0$
C
$x^{2} + y^{2} - 4 x - 4 y = 0$
D
$x^{2} + y^{2} - 4 x - 4 y - 4 = 0$

Solution

The circle meets the $x$ -axis at $A (2, 0)$ and $B (- 2, 0)$ . Let $R (h, k)$ be the intersection of $A Q$ and $B P$ .

Slope of $B R$ = slope of $B P$ : $\frac{k}{h + 2} = \frac{2 sin α}{2 cos α + 2} = tan \frac{α}{2}$ .

Slope of $A R$ = slope of $A Q$ : $\frac{k}{h - 2} = \frac{2 sin β}{2 cos β - 2} = - cot \frac{β}{2}$ .

So $tan \frac{α}{2} = \frac{k}{h + 2}$ and $tan \frac{β}{2} = \frac{2 - h}{k}$ . Given $α - β = \frac{π}{2}$ , so $\frac{α}{2} - \frac{β}{2} = \frac{π}{4}$ , hence $tan (\frac{α}{2} - \frac{β}{2}) = 1$ .

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

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

$\Rightarrow \frac{k ^{2} + h ^{2} - 4}{4 k} = 1 \Rightarrow h^{2} + k^{2} - 4 k - 4 = 0$ .

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

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