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Equation of the chord AB of the circle \mathrm{x^2+y^2=2}, passing through P(2 , 2) such that
\mathrm{\mathrm{PB} / \mathrm{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+(r \sin \theta+2)^2=2}

\mathrm{\models r^2+4(\sin \theta+\cos \theta) r+6=0 \text {, so that } \frac{P B}{P A}=\frac{r_2}{r_1}}

\mathrm{\text { Now if } r_1=a, r_2=3 a \text {, }}

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

So required chord will be y – 2  = 1 ( x –2) ⇒ y = x.

Alternative solution

\mathrm{\text { PA.PB }=P^2=2^2+2^2-2=6}         ......[1]

\mathrm{\frac{P B}{P A}=3}                                                          .......[2]

\mathrm{\text { From (1) and (2), we have PA }=\sqrt{2}, P B=3 \sqrt{2}}

\mathrm{\Rightarrow \mathrm{AB}=2 \sqrt{2} \text {. Now diameter of the circle is } 2 \sqrt{2} \text { (as radius is } \sqrt{2} \text { ) }}

Hence line passes through the centre  

⇒ y = x .

 

 

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