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

### Related Formula For external tangents from point R$R$ forming angle 2theta$2\theta$ at the center, the distance d$d$ from center to intersection point obeys cos theta = fracrd$\cos \theta = \frac{r}{d}$. ### Core Logic Given circle: x^2 + y^2 - 4x - 6y - 3 = 0$x^2 + y^2 - 4x - 6y - 3 = 0$ Center O = (2, 3)$O = (2, 3)$ Radius r = sqrt(-2)^2 + (-3)^2 - (-3) = sqrt4 + 9 + 3 = sqrt16 = 4$r = \sqrt{(-2)^2 + (-3)^2 - (-3)} = \sqrt{4 + 9 + 3} = \sqrt{16} = 4$. The tangents PQ and MN intersect at some point R(h, k)$R(h, k)$. The radius vectors OA$OA$ and OB$OB$ subtend angle angle AOB = fracpi3 = 60^circ$\angle AOB = \frac{\pi}{3} = 60^{\circ}$ at the center. The line joining O$O$ and R$R$ bisects the angle angle AOB$\angle AOB$. Thus, angle AOR = 30^circ$\angle AOR = 30^{\circ}$. ### Step 1: Apply Trigonometric Relations

In the right-angled triangle Delta AOR$\Delta AOR$, OA$OA$ is the radius (r = 4$r = 4$) and OR$OR$ is the hypotenuse. cos(30^circ) = fracOAOR = fracrOR$$\cos(30^{\circ}) = \frac{OA}{OR} = \frac{r}{OR}$$ fracsqrt32 = frac4OR Rightarrow OR = frac8sqrt3$$\frac{\sqrt{3}}{2} = \frac{4}{OR} \Rightarrow OR = \frac{8}{\sqrt{3}}$$ ### Step 2: Construct the Locus Equation The distance squared between O(2,3)$O(2,3)$ and R(h,k)$R(h,k)$ is OR^2$OR^2$: OR^2 = (h - 2)^2 + (k - 3)^2 = left(frac8sqrt3right)^2$$OR^2 = (h - 2)^2 + (k - 3)^2 = \left(\frac{8}{\sqrt{3}}\right)^2$$ (h - 2)^2 + (k - 3)^2 = frac643$$(h - 2)^2 + (k - 3)^2 = \frac{64}{3}$$ h^2 - 4h + 4 + k^2 - 6k + 9 = frac643$$h^2 - 4h + 4 + k^2 - 6k + 9 = \frac{64}{3}$$ 3(h^2 + k^2 - 4h - 6k + 13) = 64$$3(h^2 + k^2 - 4h - 6k + 13) = 64$$ 3h^2 + 3k^2 - 12h - 18k + 39 - 64 = 0$$3h^2 + 3k^2 - 12h - 18k + 39 - 64 = 0$$ 3(h^2 + k^2) - 12h - 18k - 25 = 0$$3(h^2 + k^2) - 12h - 18k - 25 = 0$$ ### Step 3: Generalize the Equation Replace (h, k)$(h, k)$ with (x, y)$(x, y)$ for the general locus: 3(x^2 + y^2) - 12x - 18y - 25 = 0$$3(x^2 + y^2) - 12x - 18y - 25 = 0$$ ### Pattern Recognition The locus of the intersection of tangents enclosing a constant angle is simply a concentric circle. Its radius expands by 1/sin(alpha/2)$1/\sin(\alpha/2)$ or 1/cos(theta)$1/\cos(\theta)$ depending on whether the angle is measured at intersection or center. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines