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  • The values of for which the common chord of the circles <img alt="\mathrm{x^2+y^2=8\:

The values of \mathrm{' a '} for which the common chord of the circles \mathrm{x^2+y^2=8\: and\: (x-a)^2+y^2=8} subtends a right angle at the origin are
 

Option: 1

\pm \sqrt{2}

 


Option: 2

\pm 2
 


Option: 3

\pm 4
 


Option: 4

None of these


Answers (1)

best_answer

Common chord is given by

\mathrm{\left(x^2+y^2-8\right)-\left((x-a)^2+y^2-8\right)=0 \Rightarrow 2 a x-a^2=0}

\mathrm{\Rightarrow \frac{2 x}{a}=1 }

Now \mathrm{ x^2+y^2-8\left(\frac{2 x}{a}\right)^2=0 } gives the combined equation of the straight lines joining the end points of this common chord and origin. If this common chord subtends a right angle at origin then \mathrm{ 1-\frac{32}{a^2}+1=0 \quad \Rightarrow a^2=16 \Rightarrow a= \pm 4 . }

Hence option 3 is correct.
 

Posted by

HARSH KANKARIA

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