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  • A line is drawn through a fixed point  to cut the circle <img alt="\mathrm{x 2

A line is drawn through a fixed point P(\alpha, \beta)  to cut the circle \mathrm{x 2+y 2=r 2} at \mathrm{A} and \mathrm{B}. Then \mathrm{P A. PB} is equal to

Option: 1

(\alpha+\beta)^{2}-r^{2}


Option: 2

\alpha^{2}+\beta^{2}-r^{2}


Option: 3

(\alpha-\beta)^{2}+r^{2}


Option: 4

None of these


Answers (1)

best_answer

The equation of any line through \mathrm{P}(\alpha, \beta) is
\mathrm{\frac{x-\alpha}{\cos \theta}=\frac{y-\beta}{\sin \theta}=k(\text { say })}

Any point on this line is \mathrm{(\alpha+k \cos \theta, \beta+k \sin \theta)}. This point lies on the given circle is \mathrm{(\alpha+\mathrm{k} \cos \theta) 2+(\beta+\mathrm{k} \sin \theta) 2=\mathrm{r} 2}

\mathrm{or \: \mathrm{k} 2+2 \mathrm{k}(\alpha \cos \theta+\beta \sin \theta)+\alpha 2+\beta 2-\mathrm{r} 2=0 \quad \ldots(i)}
which being quadratic in \mathrm{k}, gives two values of \mathrm{k} and hence the distances of two points \mathrm{A} and \mathrm{B} on the circle from the point \mathrm{P}. Let \mathrm{\mathrm{PA}=\mathrm{k} 1, \mathrm{~PB}=\mathrm{k} 2}, where \mathrm{k} 1 and \mathrm{k} 2 are the roots of (i), then

\mathrm{PA} . \mathrm{PB}=\mathrm{k} 1 \mathrm{k} 2=\alpha 2+\beta 2-\mathrm{r} 2

Posted by

shivangi.bhatnagar

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