JEE Main 2026 — 21 January, Morning Shift — Question 15

$$\text{Given circle: } x^{2}+y^{2}-4x-6y-3=0$$

Completing squares,

$$(x-2)^2+(y-3)^2=16$$

Hence, the centre is

$$O(2,3), \quad r=4.$$

Let the tangents $PQ$ and $MN$ touch the circle at points $A$ and $B$ respectively, and let their point of intersection be $P(x,y)$. Then

$$OA \perp PA, \qquad OB \perp PB.$$

Given

$$\angle AOB=\frac{\pi}{3}.$$

In quadrilateral $OAPB$,

$$\angle APB=\pi-\angle AOB=\pi-\frac{\pi}{3}=\frac{2\pi}{3}.$$

Thus, the angle between the tangents drawn from $P$ to the circle is constant. Hence, the locus of $P$ is a circle concentric with the given circle.

Now,

$$\sin\frac{\angle AOB}{2}=\sin\frac{\pi}{6}=\frac{1}{2}.$$

For tangents from $P$,

$$OP=\frac{r}{\sin(\angle AOB/2)} =\frac{4}{1/2}=8.$$

Therefore, the locus of $P$ is the circle

$$(x-2)^2+(y-3)^2=64.$$

Expanding,

$$x^2+y^2-4x-6y-51=0.$$

Multiplying throughout by $3$,

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