JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Circle

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

∠ AOB = π /3

. Then the locus of the point of intersection of the lines PQ and MN is:

Answer: (D)

Solution

The circle has centre

O (2, 3)

and radius

r = 4 + 9 + 3 = 4

. If the tangents from a point

R (h, k)

touch at

A

and

B

with

∠ A O B = \frac{π}{3}

, then each radius makes

\frac{π}{6}

with

O R

; from the right angle at the point of contact,

cos \frac{π}{6} = \frac{r}{O R}

, so

O R = \frac{8}{3}

and

O R^{2} = \frac{64}{3}

. Thus

(h - 2)^{2} + (k - 3)^{2} = \frac{64}{3}

, i.e.

h^{2} + k^{2} - 4 h - 6 k + 13 = \frac{64}{3}

; multiplying by

3

, the locus is

3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option