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The equation of the circle which passes through \mathrm{(2 a, 0)} and has the radical axis \mathrm{2 x-a=0} with the circle \mathrm{x^2+y^2=a^2} is
 

Option: 1

\mathrm{ x^2+y^2-2 a x=0}


Option: 2

\mathrm{ x^2+y^2+2 a x=0}


Option: 3

\mathrm{ x^2+y^2+3 a x=0}


Option: 4

\mathrm{ x^2+y^2-3 a x=0 }


Answers (1)

best_answer

 The circle is \mathrm{\left(x^2+y^2-a^2\right)+\lambda(2 x-a)=0}

\therefore  it passes through \mathrm{(2 a, 0)}, then we have, \mathrm{\lambda=-a}

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Ajit Kumar Dubey

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