Question

$L_{1}$ and $L_{2}$ are two common targets to two circles. If $L_{1}$ touches the two circles at $A (1,1)$ and $B (0,1)$ and $L_{2}$ touches the two circles at $C\left(\frac{3}{5}, \frac{4}{5}\right), D\left(-\frac{1}{5}, \frac{7}{5}\right)$, then the equation of the radical axis of the two circles is

• A. $2 x-6 y=7$
• B. $2 x+y+7=0$
• C. $2 x+6 y=7$
• D. $x=y$

Answer 1

Answer: (C)

As radical axis of the two circles bisects the all common tangents of the circles.
Now, the mid-point of $A(1,1)$ and $B(0,1)$ is $M\left(\frac{1}{2}, 1\right)$ and mid-point of $C\left(\frac{3}{5}, \frac{4}{5}\right)$ and $D\left(-\frac{1}{5}, \frac{7}{5}\right)$ is $N\left(\frac{1}{5}, \frac{11}{5}\right)$
$\therefore $ Equation of required radical axis $M N$ is
$y-1=\frac{\frac{11}{10}-1}{\frac{1}{5}-\frac{1}{2}}\left(x-\frac{1}{2}\right)$
$\Rightarrow 3(y-1)=-x+\frac{1}{2}$
$\Rightarrow 2 x+6 y=7$