From the point $O$ $(2, 3)$ , tangents $O P, O Q$ are drawn to circle $x^{2} + y^{2} = 1$ . Equation to the linejoining the midpoint of $O P$ and $O Q$ is $\dots \dots$

2 x + 3 y = 7

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3 x + 2 y = 7

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x + 2 y = 3

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2 x + y = 3

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Solution

The correct option is A $2 x + 3 y = 7$
$e q^{n}$ of chord of contact from $O (2, 3) t o x^{2} + y^{2} = 1$
$2 x + 3 y = 1$
So, $x = \frac{1 - 3 y}{2}$
${(\frac{1 - 3 y}{2})}^{2} + y^{2} = 1$
$9 y^{2} - 6 y + 1 + 4 y^{2} = 1$
$13 y^{2} - 6 y = 0$
$y = 0 o r y = \frac{6}{13}$
So, $x = \frac{1}{2} o r x = \frac{- 5}{26}$
So, $p \equiv (\frac{1}{2}) a n d (\frac{- 5}{26}, \frac{6}{13})$
So, mid point of of is $m (\frac{5}{4}, \frac{3}{4})$
and mid - point of $O i s N (\frac{47}{52}, \frac{45}{26})$
So, $e q^{n}$ of line joining M N,
$\Rightarrow 3 y + 2 x = 7$

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