If the curve $\frac{x^{2}}{4} + y^{2} = 1$ and $\frac{x^{2}}{a^{2}} + y^{2} = 1$ for suitable values of $^{'} a^{'}$ cut on four concyclic points, the equation of the circle passing through these points is

x^{2} + y^{2} = 2

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x^{2} + y^{2} = 1

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x^{2} + y^{2} = 4

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x^{2} + y^{2} = 3

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Solution

The correct option is B $x^{2} + y^{2} = 1$
Let the equation of the circle be
$S_{1} + λ S_{2} = 0$
$\Rightarrow (\frac{x^{2}}{4} + y^{2} - 1) + λ (\frac{x^{2}}{a^{2}} + y^{2} - 1) = 0$
$\Rightarrow x^{2} (\frac{a^{2} + 4 λ}{4 a^{2} (1 + λ)}) + y^{2} = 1$
For the above equation to be a circle ;
$(\frac{a^{2} + 4 λ}{4 a^{2} (1 + λ)}) = 1$
Therefore equation of required circle is
$x^{2} + y^{2} = 1$

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