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
B
3
C
4
D
2

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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