CircleMedium
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:
Options
A.$3\left( x^{2} + y^{2} \right) - 18x - 12y + 25 = 0$
B.$x^{2} + y^{2} - 12x - 18y - 25 = 0$
C.$x^{2} + y^{2} - 18x - 12y - 25 = 0$
D.$3\left( x^{2} + y^{2} \right) - 12x - 18y - 25 = 0$
Solution
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
$${{OR}^{2} = (h - 2)^{2} + (k - 3)^{2} }{\Rightarrow 3\left( x^{2} + y^{2} \right) - 12x - 18y - 25 = 0}$$
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