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  • Let  be any point on <img alt="\mathrm{x^{2}+y^{2}=16, Q}" src="https://entrancecorner.oncod

Let \mathrm{P} be any point on \mathrm{x^{2}+y^{2}=16, Q} a point on line \mathrm{7 x+y+3=0} and the line \mathrm{x-y+1=0} is perpendicular bisector of \mathrm{P Q}. Then \mathrm{P } is 

Option: 1

(2,1)


Option: 2

(1,2)


Option: 3

(1,-2)


Option: 4

None of these


Answers (1)

best_answer


Let \mathrm{R(\alpha, \alpha+1)} be \mathrm{\perp} bisector of \mathrm{P Q}.
Then equation of \mathrm{P Q} is
\mathrm{\frac{x-\alpha}{\cos \theta}=\frac{y-(\alpha+1)}{\sin \theta}}\quad \cdots(1)

Its slope is \mathrm{\tan \theta=-1}
\mathrm{\sin \theta=\frac{1}{\sqrt{2}} \cos \theta=-\frac{1}{\sqrt{2}}} so (1) becomes.

\mathrm{\frac{x-\alpha}{-\frac{1}{\sqrt{2}}}=\frac{y-(\alpha+1)}{\frac{1}{\sqrt{2}}}=r( for \: point \: Q)}
\mathrm{x=\alpha^{-\frac{r}{\sqrt{2}}}, y=\alpha+1+\frac{r}{\sqrt{2}}}
Putting in straight line
\mathrm{7 x+y+3=0}
\mathrm{ 8 \alpha+4-\frac{6 r}{\sqrt{2}}=0 }
\mathrm{ \frac{6 r}{\sqrt{2}}=8 \alpha+4 }
\mathrm{ r=\frac{(8 \alpha+4) \sqrt{2}}{6} }

For point 'P':-
\mathrm{ \frac{x-\alpha}{-\frac{1}{\sqrt{2}}}=\frac{y-(\alpha+1)}{\frac{1}{\sqrt{2}}}=-r=-\frac{(8 \alpha+4) \sqrt{2}}{6} }
\mathrm{ x-\alpha=\frac{8 \alpha+4}{6} }
\mathrm{ x=\alpha+\frac{8 \alpha+4}{6}=\frac{14 \alpha+4}{6} }
\mathrm{ y=\alpha+1-\frac{(8 \alpha+4)}{6}=\frac{6 \alpha+6-8 \alpha-4}{6}=\frac{-2 \alpha+2}{6} }

Putting in circle
\mathrm{ x^{2}+y^{2}=16 }, we get:
\mathrm{ \left(\frac{14 \alpha+4}{6}\right)^{2}+\left(\frac{-2 \alpha+2}{6}\right)^{2}=16 }
\mathrm{ 200 \alpha^{2}+104 \alpha-556=0 }
\mathrm{ \Rightarrow 50 \alpha^{2}+26 \alpha-139=0 }

We will get a point which is not in the alternatives.

 

Posted by

Shailly goel

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