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The equation of the circle, which passes through the point \mathrm{(2 a, 0) } and whose radical axis is \mathrm{x=\frac{a}{2}} with respect to the circle

\mathrm{x^2+y^2=a^2}, will be

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+2 a y=0}


Option: 4

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


Answers (1)

best_answer

Equation of radical axis is \mathrm{x=\frac{a}{2} \Rightarrow 2 x-a=0}

Equation of required circle is \mathrm{x^2+y^2-a^2+\lambda(2 x-a)=0}

  It is passes through the point \mathrm{(2 a, 0), \quad \therefore \quad 4 a^2-a^2+\lambda(4 a-a)=0 \Rightarrow \lambda=-a}

\mathrm{\therefore \text { Equation of circle is } x^2+y^2-a^2-2 a x+a^2=0 \Rightarrow x^2+y^2-2 a x=0}

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manish painkra

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