L_1 and L_2 are two common tangents to two ci | vedclass

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(C) The radical axis of two circles bisects all common tangents of the circles.First,find the midpoint $M$ of $A(1, 1)$ and $B(0, 1)$:$M = \left(\frac

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$2x + 6y = 7$

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(C) The radical axis of two circles bisects all common tangents of the circles.First,find the midpoint $M$ of $A(1, 1)$ and $B(0, 1)$:$M = \left(\frac{1+0}{2}, \frac{1+1}{2}\right) = \left(\frac{1}{2}, 1\right)$.Next,find the midpoint $N$ of $C\left(\frac{3}{5}, \frac{4}{5}\right)$ and $D\left(-\frac{1}{5}, \frac{7}{5}\right)$:$N = \left(\frac{3/5 - 1/5}{2}, \frac{4/5 + 7/5}{2}\right) = \left(\frac{2/5}{2}, \frac{11/5}{2}\right) = \left(\frac{1}{5}, \frac{11}{10}\right)$.The radical axis passes through $M$ and $N$. The slope $m$ of the line $MN$ is:$m = \frac{11/10 - 1}{1/5 - 1/2} = \frac{1/10}{-3/10} = -\frac{1}{3}$.The equation of the line $MN$ is:$y - 1 = -\frac{1}{3}\left(x - \frac{1}{2}\right)$$3(y - 1) = -x + \frac{1}{2}$$3y - 3 = -x + 0.5$$x + 3y = 3.5$$2x + 6y = 7$.

$L_1$ 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 $C\left(\frac{3}{5}, \frac{4}{5}\right)$ and $D\left(-\frac{1}{5}, \frac{7}{5}\right)$,then the equation of the radical axis of the two circles is

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