The line `x + 2y = a` intersects the circle `x^2 + y^2 = 4` at two distinct points `A and B` Another line `12x - 6y - 41 = 0` intersects the circle `x^2 + y^2 - 4x - 2y + 1 = 0` at two `C and D`. The value of 'a' for which the points `A,B,C and D` are concyclic -

Options

A. 1
B. 3
C. 4
D. 2

Answer

4

Explanation

The given circles are
`C_(1) : x^(2)+y^(2) - 4 =0 `
and `C_(2) : x^(2)+y^(2) - 4x-2y +1=0`
The given lines are
`L_(1) : x+2y +a=0`
and `L_(2) : 12 x - 6y -41 =0`
If A,B,C, and D are concyclic , then
`PA. PB = PC. PD`
or `PT^(2) = PT^('2)`
Therefore, point P will lie on the radical axis of circles `C_(1)` and `C_(2) ` i.e., `L_(3) : 4x+2y-5=0`
Now, lines `L_(1), L_(2)`, and `L_(3)` are concurrent. Therefore,
`|{:(4,2,-5),(1,2,a),(12,-6,-41):}|0` or `a=2`
Also , for this value of a, the line meets circle (1) at two distinct points.