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

Option: 1

\mathrm{x=3 y}


Option: 2

\mathrm{x=y}


Option: 3

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


Option: 4

none of these


Answers (1)

best_answer

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

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

\mathrm{\begin{aligned} \Rightarrow \quad & r^2+4(\sin \theta+\cos \theta) r+6=0 \\ & \frac{P B}{P A}=\frac{r_2}{r_1}, \text { now if } r_1=\alpha, r_2=3 \alpha, \end{aligned}}

then  \mathrm{4 \alpha=-4(\sin \theta+\cos \theta), 3 \alpha^2=6 \Rightarrow \sin 2 \theta=1 \Rightarrow \theta=\pi / 4}

So required chord will be  \mathrm{y-2=1(x-2) \Rightarrow y=x}.

Alternative solution:

\mathrm{\text { PA. } \mathrm{PB}=\mathrm{PT}^2=2^2-2=6}                                                  ...........(1)

\mathrm{\frac{\mathrm{PB}}{\mathrm{PA}}=3}                                                                                              ............(2)

From, (1) and (2) we have \mathrm{\mathrm{PA}=\sqrt{2}, \mathrm{~PB}=3 \sqrt{2}}

\mathrm{\Rightarrow \mathrm{AB}=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 y=x}.

Hence (B) is the correct answer.

Posted by

Kuldeep Maurya

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