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Let PQ and MN be two straight lines touching the circle... | null (Mathematics) | DivineJEE

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Mathematics

JEE Mains 2026 21st January Shift 1

Mathematics

Let PQ and MN be two straight lines touching the circle

x^{2}+y^{2}-4 x-6 y-3=0

at the points A and B respectively. Let O be the centre of the circle and

\angle \mathrm{AOB}=\pi / 3

. Then the locus of the point of intersection of the lines PQ and MN is:

A

3\left(x^{2}+y^{2} ight)-18 x-12 y+25=0

B

x^{2}+y^{2}-12 x-18 y-25=0

C

x^{2}+y^{2}-18 x-12 y-25=0

D

3\left(x^{2}+y^{2} ight)-12 x-18 y-25=0

Explanation

Given circle

\\

x^{2}+y^{2}-4 x-6 y-3=0\\

C(2,3) \& r=4\\

\cos 30^{\circ}=\frac{\mathrm{r}}{\mathrm{OR}}=\frac{4}{\mathrm{OR}}\\

\Rightarrow \mathrm{OR}=\frac{8}{\sqrt{3}}\\

Now

\\ \mathrm{OR}^{2}=(\mathrm{h}-2)^{2}+(\mathrm{k}-3)^{2}\\

\Rightarrow 3\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)-12 \mathrm{x}-18 \mathrm{y}-25=0

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