JEE Mathematics 2026 Question with Solution

Given: The circle is $x^2+y^2-4x-6y-3=0$. The tangents intersect at a point $P(h,k)$, and $\angle AOB=\pi/3$.

Find: The locus of the point of intersection of the tangents.

For the circle $x^2+y^2+2gx+2fy+c=0$, the centre is $(-g,-f)$ and the radius is $\sqrt{g^2+f^2-c}$.

So the centre is $O(2,3)$ and the radius is

$$r=\sqrt{(-2)^2+(-3)^2-(-3)}=\sqrt{4+9+3}=4$$

Since $OA$ and $OB$ are radii to the points of contact, they are perpendicular to the tangents. Thus, in quadrilateral $OAPB$, triangles involving $OA, OP, PA$ and $OB, OP, PB$ are right-angled at the points of contact.

Also, $OP$ bisects $\angle AOB$, so

$$\angle AOP=\frac{1}{2}\angle AOB=\frac{1}{2}\cdot \frac{\pi}{3}=\frac{\pi}{6}$$

Now in right triangle $OAP$,

$$\cos \angle AOP = \frac{OA}{OP}$$

So,

$$\cos 30^\circ = \frac{4}{OP}$$ $$\frac{\sqrt{3}}{2}=\frac{4}{OP}$$ $$OP=\frac{8}{\sqrt{3}}$$

Hence the distance of $P(h,k)$ from $O(2,3)$ is constant, so the locus is a circle:

$$(h-2)^2+(k-3)^2=\left(\frac{8}{\sqrt{3}}\right)^2=\frac{64}{3}$$

Replacing $h,k$ by $x,y$,

$$(x-2)^2+(y-3)^2=\frac{64}{3}$$

Expanding,

$$3[(x^2-4x+4)+(y^2-6y+9)]=64$$ $$3(x^2+y^2-4x-6y+13)=64$$ $$3x^2+3y^2-12x-18y+39=64$$ $$3x^2+3y^2-12x-18y-25=0$$

That is,

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

Therefore, the correct option is C.

Given: Two tangents from an external point intersect the given circle at points of contact $A$ and $B$, with centre $O(2,3)$ and radius $4$.

Find: The locus of the external point.

The key idea is that for tangents drawn from an external point, the line joining the external point to the centre bisects the angle between the radii to the points of contact. Since $\angle AOB=60^\circ$, the angle in triangle $OAP$ at the centre is $30^\circ$.

Using the right triangle $OAP$,

$$\cos 30^\circ = \frac{OA}{OP}=\frac{4}{OP}$$

This fixes $OP$ at a constant value $\frac{8}{\sqrt{3}}$.

Therefore, the point of intersection of tangents always remains at a constant distance from the fixed point $O(2,3)$. Hence its locus is a circle with centre $O(2,3)$ and radius $\frac{8}{\sqrt{3}}$.

Its equation is

$$(x-2)^2+(y-3)^2=\frac{64}{3}$$

which simplifies to

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

So the required locus is the circle given in option C.