The line `x+2y+a=0` intersects the circle `x^(2)+y^(2)-4=0` at two distinct points A and B. Another line `12x-6y-41=0` intersects the circle `x^(2)+y^(2)-2y+1=0` at two distinct point C and D.
The equation of the circle passing through the points A,B,D, and D is

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.