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

Option: 1

x=3 y


Option: 2

x=y


Option: 3

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


Option: 4

none of these


Answers (1)

best_answer

Key concept: Using the concept of parametric equation of any a line.
Any line passing through (2,2) will be of the form \mathrm{\frac{y-2}{\sin \theta}=\frac{x-2}{\cos \theta}=r}  If this line cuts the circle \mathrm{x^{2}+y^{2}=2}, then \mathrm{(r \cos \theta+2)^{2}+(r \sin \theta+2)^{2}=2\, \Rightarrow r^{2}+4(\sin \theta+\cos \theta) r+6=0}.This is a quadratic equation in \mathrm{r}, roots of this equation will represent distance PB and PA. Let the roots be \mathrm{r_{1}} and \mathrm{r_{2}}.

Then \mathrm{\frac{P B}{P A}=\frac{r_{2}}{r_{1}}, now \: if \: r_{1}=\alpha, r_{2}=3 \alpha},
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:
Key concept: Using the basic property of a circle.

\mathrm{P A \cdot P B=P T^{2}=2^{2}+2^{2}-2=6}\quad \ldots(1)
\mathrm{\frac{\mathrm{PB}}{\mathrm{PA}}=3}\quad \ldots(2)

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

Posted by

Shailly goel

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