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Question

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


A
1
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B
3
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C
4
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D
2
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Solution

The correct option is C $$2$$
Let lines $$x+2y+a = 0$$ and $$12x-2y+1=0$$intersects at $$p$$, then $$PA.PB = PT^2$$ and $$PC.PD = PT'^2$$, where $$t$$ and $$T'$$ are the point on the respective circles.
A, B, C, and D are concyclic.
$$PA.PB = PC.PD = PT^2 = PT'^2$$
Hence, point $$P$$ will lie on the radical axis of both the circles. 
Now equation of the radical axis is
$$4x+2y-5=0$$
Since, radical axis and lines $$x+2y +a=0$$ and $$12x-6y-41=0$$ are concurrent at $$P$$, we have
$$\begin{vmatrix} 4&2&-5\\1&2& a\\12&-6&-41\end{vmatrix} = 0 \Rightarrow a =2$$

Mathematics

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