If the curves `(x^2)/4+y^2=1` and `(x^2)/(a^2)+y^2=1` for a suitable value of `a` cut on four concyclic points, the equation of the circle passing through these four points is `x^2+y^2=2` (b) `x^2+y^2=1` `x^2+y^2=4` (d) none of these

Options

A. `x^(2)+y^(2)=2`
B. `x^(2)+y^(2)=1`
C. `x^(2)+y^(2) =4`
D. none of these

Answer

B

Explanation

Let equation of the circle be `((x^(2))/(4)+y^(2)-1) + lambda ((x^(2))/(a^(2))+y^(2)-1) = 0`
`rArr x^(2) ((1)/(4)+(lambda)/(a^(2))) +y^(2) (1+lambda) =1 +lambda`
`rArr x^(2) ((a^(2)+4 lambda)/(4a^(2))) +y^(2) (1+lambda) =1 +lambda`
`rArr x^(2) ((a^(2)+4 lambda)/(4a^(2)(1+lambda))) +y^(2) = 1`
Clearly, the circle is `x^(2)+y^(2) =1`.