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The equation of a circle which passes through (2 \mathrm{a}, 0) and whose radical axis in relation to the circle \mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2} is \mathrm{x}=\mathrm{a} / 2, is

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

The required circle is
\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{a}^{2}+\lambda\left(x-\frac{a}{2}\right)=0 \text { (using } \mathrm{S}+\lambda \mathrm{L}=0 \text { ) }

This passes through (2 \mathrm{a}, 0)
4 \mathrm{a}^{2}-\mathrm{a}^{2}+\left(\frac{3 a}{2}\right) \lambda=0 \Rightarrow \lambda=-2 a
Hence, the required circle is
\begin{aligned} & x^{2}+y^{2}-a^{2}-2 a\left(x-\frac{a}{2}\right)=0 \\ & x^{2}+y^{2}-a^{2}-2 a x+a^{2}=0 \Rightarrow x^{2}+y^{2}-2 a x=0 \end{aligned}

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Irshad Anwar

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