Equation of chord $$mathrm{AB}$$ of circle $$x^2+y^2=2$$ passing through $$mathrm{mathrm{P}(2,2)}$$ such that $$mathrm{PB} / mathrm{PA}=3$$ , is given by

Equation of chord of circle <img alt="\mathrm{\mathrm{x} 2+\mathrm{y} 2=2}" src="https://entranceco

Equation of chord $\mathrm{AB}$ of circle $\mathrm{\mathrm{x} 2+\mathrm{y} 2=2}$ passing through $\mathrm{\mathrm{P}(2,2)}$ such that $\mathrm{PB} / \mathrm{PA}=3$ , is given by
Option: 1
$\mathrm{x=3 y}$

Option: 2
$\mathrm{\quad x=y}$

Option: 3
$\mathrm{y-2=\sqrt{3}(x-2)}$

Option: 4
none of these

Answers (1)

Any line passing through $(2,2)$ will be of the form $\frac{y-2}{\sin \theta}=\frac{x-2}{\cos \theta}=r$

When this line cuts the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=2,(2+\mathrm{y} \cos \theta)^{2}+(2+\mathrm{y} \sin \theta)^{2}=2$
$\mathrm{\Rightarrow r^{2}+4(\sin \theta+\cos \theta) r+6=0}$

$\mathrm{\frac{P B}{P A}=\frac{r_{2}}{r_{1}}}$ , now if $\mathrm{r_{1}= a, r_{2}= 3a}$ ,

$\mathrm{then \: 4 \mathrm{a}=-4(\sin \theta+\cos \theta), 3 a^{2}=6}$

$\mathrm{\Rightarrow \sin 2 \theta=1}$
$\mathrm{\Rightarrow \theta=\pi / 4}$
So required chord will be $\mathrm{\mathrm{y}-2=1(\mathrm{x}-2)}$
$\mathrm{\mathrm{y}=\mathrm{x}}$

Alternative Solution
$\mathrm{\mathrm{PA} . \mathrm{PB}=\mathrm{PT}^{2}=2^{2}=2^{2}-2=6 \quad \ldots(i)}$

$\mathrm{\frac{P B}{P A}=3 \quad \ldots(ii)}$
From (i) and (ii), we have $\mathrm{P A=\sqrt{2}, P B=3 \sqrt{2}}$
$\mathrm{\Rightarrow \quad A B=2 \sqrt{2}}$ . Now diameter of the circle is $\mathrm{2 \sqrt{2}}$ (as radius is $\mathrm{\sqrt{2}}$

Hence line passes through the centre $\mathrm{\Rightarrow \mathrm{y}=\mathrm{x}}$ .
Hence (b) is the correct answer.