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  • Equation of chord AB of circle   passing through P(2, 2)

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

Option: 1

\mathrm{x=3 y}


Option: 2

\mathrm{x=y}


Option: 3

\mathrm{y}-2=\sqrt{3}(\mathrm{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{PA. \mathrm{PB}=\mathrm{PT}^2=2^2-2=6 \frac{\mathrm{PB}}{\mathrm{PA}}=3 } ----(i) , (ii)

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 \Rightarrow y=x
Hence (B) is the correct answer.

Posted by

Ritika Harsh

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