Let PQ and MN be two straight lines touching the circle $x^{2} + y^{2} - 4x - 6y - 3 = 0$ at the poi

Question

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:

Accepted Answer

Given circle$${x^{2} + y^{2} - 4x - 6y - 3 = 0 }{C(2,3)\& r = 4 }{cos30^{\circ} = \frac{r}{OR} = \frac{4}{OR} }{\Rightarrow OR = \frac{8}{\sqrt{3}} }$$Now$${{OR}^{2} = (h - 2)^{2} + (k - 3)^{2} }{\Rightarrow 3\left( x^{2} + y^{2} \right) - 12x - 18y - 25 = 0}$$

Question

Options

Solution

More Straight Line Questions