Animated Solution for Mathematics - Circles: Let $PQ$ and $MN$ be two straight lines touching the circle $x^2+y^2-4x-6y-3=0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $\angle AOB=\pi /3$. Then the locus of the point of intersection of the lines $PQ$ and $MN$ is :

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the elegant dance between a circle and the lines that kiss its boundary. Imagine you are standing at the center of a circle, watching two lines, $PQ$ and $MN$, converge from the distance to touch the circle at points $A$ and $B$.

The problem asks us to find the locus of their intersection, $P$. This is a classic JEE Advanced setup—it tests your ability to see the hidden symmetry in a seemingly complex algebraic expression.

Before we can dance, we must know the stage. We are given the circle equation $x^2+y^2-4x-6y-3=0$. To understand its heart, we compare it to the general form $x^2+y^2+2gx+2fy+c=0$.

By matching coefficients, we find $g=-2$, $f=-3$, and $c=-3$. The center $O$ is at $(-g,-f)$, which gives us the coordinates $(2,3)$.

The radius $r$ is calculated as:

Our circle is centered at $(2,3)$ with a radius of $4$.

Now, visualize the points of contact $A$ and $B$. We know that the radius is always perpendicular to the tangent at the point of contact. This means $\angle OAP=90^{\circ }$ and $\angle OBP=90^{\circ }$.

The line connecting the center $O$ to the external intersection point $P$ is a powerful tool—it acts as an axis of symmetry, bisecting the angle $\angle AOB$. Since we are given $\angle AOB=60^{\circ }$, the angle $\angle AOP$ must be exactly $30^{\circ }$.

Focus your attention on the right-angled triangle $△OAP$. We know the base $OA$ is the radius, $4$. We know the angle $\angle AOP=30^{\circ }$. We need the hypotenuse $OP$.

Using trigonometry:

Substituting our values:

This is the distance from the center $O$ to the point $P$. Because this distance is constant, the point $P$ must trace a circle centered at $O$ with radius $R=\frac{8}{\sqrt{3}}$.

Now, we translate this geometric truth into algebra. The locus of $P(x,y)$ is the set of points at a distance $\frac{8}{\sqrt{3}}$ from $(2,3)$. The equation is:

Expanding this, we get:

Simplifying the constants:

To clear the fraction, multiply the entire equation by $3$:

Finally, subtracting $64$ from both sides, we arrive at the final locus:

This is not just an equation; it is the path of a point moving in perfect harmony with the circle. You have successfully navigated the geometry and the algebra. Keep this clarity, and you will conquer any problem the JEE throws at you.