JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

Let PQ and MN be two straight lines touching the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ at the points A and B respectively. Let O be the centre of the circle and $∠ A O B = π /3$ . Then the locus of the point of intersection of the lines PQ and MN is:

A
$3 (x^{2} + y^{2}) - 18 x - 12 y + 25 = 0$
B
$x^{2} + y^{2} - 12 x - 18 y - 25 = 0$
C
$x^{2} + y^{2} - 18 x - 12 y - 25 = 0$
D
$3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0$

Solution

The circle has centre $O = (2, 3)$ and radius $r = 4 + 9 + 3 = 4$ .

If $R$ is the intersection of the two tangents, the triangle $O A R$ has a right angle at $A$ with $∠ A O R = 30°$ (half of $∠ A O B$ ). So $cos 30° = \frac{r}{∣ O R ∣} \Rightarrow ∣ O R ∣ = \frac{8}{3}$ , i.e. $∣ O R ∣^{2} = \frac{64}{3}$ .

Locus of $R = (x, y)$ : $(x - 2)^{2} + (y - 3)^{2} = \frac{64}{3}$ , which simplifies to $3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0$ .

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