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$L_1$ and $L_2$ are two common tangents 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
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The correct answer is:
$2 x+6 y=7$
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
$y-1=\frac{\frac{11}{10}-1}{\frac{1}{5}-\frac{1}{2}}\left(x-\frac{1}{2}\right) \Rightarrow 3(y-1)=-x+\frac{1}{2}$
$\Rightarrow \quad 2 x+6 y=7$
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
$y-1=\frac{\frac{11}{10}-1}{\frac{1}{5}-\frac{1}{2}}\left(x-\frac{1}{2}\right) \Rightarrow 3(y-1)=-x+\frac{1}{2}$
$\Rightarrow \quad 2 x+6 y=7$
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