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. $2$

Answer 1

Answer: (D)

$2$

Let lines $x+2y+a=0$ and $12x-2y+1=0$intersects at $p$, then $PA.PB=P{T}^2$ and $PC.PD=P{T}^{\prime 2}$, where $t$ and $T^{\prime }$ are the point on the respective circles.

A, B, C, and D are concyclic.

$PA.PB=PC.PD=P{T}^2=P{T}^{\prime 2}$

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

$\left|\begin{array}{ccc}4& 2& -5\\ 1& 2& a\\ 12& -6& -41\end{array}\right|=0\Rightarrow a=2$