Step-by-Step Calculation
1. Analyze the Given Circle:
The equation is $x^2 + y^2 – 4x – 6y – 3 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$:
$g = -2, f = -3, c = -3$.
Centre $O = (-g, -f) = (2, 3)$.
Radius $r = \sqrt{g^2 + f^2 – c} = \sqrt{4 + 9 – (-3)} = \sqrt{16} = 4$.
2. Geometry of the Intersection:
Let the point of intersection of the tangents PQ and MN be $P(h, k)$.
In $\triangle OAP$ (where A is the point of contact), $\angle OAP = 90^\circ$ because the tangent is perpendicular to the radius.
The line $OP$ bisects $\angle AOB$.
Given $\angle AOB = \pi/3$, so $\angle AOP = \frac{\angle AOB}{2} = \frac{\pi/3}{2} = \pi/6$.
3. Distance Calculation:
In right-angled $\triangle OAP$:
$\cos(\angle AOP) = \frac{OA}{OP}$
$\cos(\pi/6) = \frac{r}{OP} \implies \frac{\sqrt{3}}{2} = \frac{4}{OP}$
$OP = \frac{8}{\sqrt{3}}$.
4. Formulating the Locus:
The distance squared is $OP^2 = \frac{64}{3}$.
Using the distance formula for $P(h, k)$ and $O(2, 3)$:
$(h-2)^2 + (k-3)^2 = \frac{64}{3}$
$3(h^2 – 4h + 4 + k^2 – 6k + 9) = 64$
$3(h^2 + k^2 – 4h – 6k + 13) = 64$
$3h^2 + 3k^2 – 12h – 18k + 39 – 64 = 0$
$3h^2 + 3k^2 – 12h – 18k – 25 = 0$.
Replacing $(h, k)$ with $(x, y)$, the locus is:
3(x^2 + y^2) – 12x – 18y – 25 = 0.
Advanced Analytical Theory
Foundations of Circle Geometry in Coordinate Systems
The study of circles in the Cartesian plane is a cornerstone of Coordinate Geometry. A circle is defined as the locus of a point that moves such that its distance from a fixed point (the center) remains constant. When we transition from the basic equation $(x-h)^2 + (y-k)^2 = r^2$ to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, we open the door to sophisticated transformations and power-of-a-point theorems. In JEE Advanced preparations, the ability to decompose the general equation instantly into its center and radius components is paramount.
The Nature of Tangents and Normals
A tangent to a circle is a limiting case of a secant where the two points of intersection coincide. Mathematically, for a point $P(x_1, y_1)$ on the circle, the tangent is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$. However, when the point of intersection is external, as in this problem, the tangents represent a pair of lines whose slopes are determined by the quadratic constraint that their distance from the center must equal the radius.
Locus and Fixed Angular Constraints
When two tangents intersect at a point such that the angle between them (or the angle between the radii to the points of contact) is constant, the resulting locus of the intersection point is always a concentric circle. This is due to the inherent symmetry of the circle. If the angle between the tangents is $2\phi$, the distance from the center to the intersection point $L$ is constant at $r \cdot \csc(\phi)$. In this specific problem, we were given the angle between the radii ($\angle AOB = \pi/3$), which implies the angle between the tangents is $\pi – \pi/3 = 2\pi/3$. The geometry remains rigid even as the tangents rotate around the circle, tracing a larger circular path.
Trigonometric Interplay in Geometry
The use of $\cos(\pi/6)$ in the solution highlights the intersection of trigonometry and geometry. In competitive exams, specific angles like $\pi/6, \pi/4, \pi/3$ are chosen because their trigonometric ratios are algebraic. The relationship $OP = r / \cos(\theta/2)$ is a universal shortcut for such locus problems. It is vital to distinguish whether the given angle is between the tangents or between the radii, as they are supplementary.
Parametric and Vector Approaches
While the algebraic approach is standard, one could also view this through a parametric lens. Let the points of contact be $A(2+4\cos\alpha, 3+4\sin\alpha)$ and $B(2+4\cos(\alpha+\pi/3), 3+4\sin(\alpha+\pi/3))$. The intersection of tangents at these points involves solving a system that eventually leads to the same concentric circle equation. Vectorially, the dot product of the radial vectors $\vec{OA} \cdot \vec{OB}$ would involve $\cos(\pi/3)$, providing an alternative path to proving the fixed distance of the intersection point from the origin.
Computational Efficiency in JEE
In a high-pressure environment, students should recognize that if the answer choices are all circles, and the locus is based on a fixed angle to a circle centered at $(2,3)$, the resulting equation must have the same $-12x$ and $-18y$ coefficients (or a multiple thereof) as the original circle’s $-4x$ and $-6y$. This “symmetry check” allows for rapid elimination of incorrect options, focusing only on the constant term.
… (This theoretical discussion continues into the properties of Director Circles, where the angle is fixed at 90 degrees, and the Pole-Polar relationship where the intersection of tangents is the pole of the chord of contact) …