Equation of chord $A B$ of circle $x^{2} + y^{2} = 2$ passing through $P (2, 2)$ such that $\frac{P B}{P A} = 3$ , is given by

x = 3 y

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x = y

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y - 2 = \sqrt{3} (x - 2)

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None of these

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Solution

The correct option is B $x = y$
Any line passing through $(2, 2)$ will be of the form $\frac{y - 2}{sin θ} = \frac{x - 2}{cos θ} = r$

When this line cuts the circle $x^{2} + y^{2} = 2, {(r cos θ + 2)}^{2} + {(r sin θ + 2)}^{2} = 2$

$\Rightarrow r^{2} {cos}^{2} θ + 4 + 4 r cos θ + r^{2} {sin}^{2} θ + 4 + 4 r sin θ = 2$

$\Rightarrow r^{2} ({cos}^{2} θ + {sin}^{2} θ) + 4 r (cos θ + sin θ) + 6 = 0$

$\frac{P B}{P A} = \frac{r_{2}}{r_{1}},$

Now if $r_{1} = α, r_{2} = 3 α$ then

$4 α = - 4 (cos θ + sin θ)$

$3 α^{2} = 6 \Rightarrow sin 2 θ = 1 \Rightarrow θ = \frac{π}{4}$

So, required chord will be $y - 2 = 1 (x - 2)$
$\Rightarrow y = x$

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