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The perpendicular distance from the point $(1,2)$ to common chord of the circles $x^2+y^2-2 x+4 y-4=0$ and $x^2+y^2+4 x-6 y-3=0$ is........ units.
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Verified Answer
The correct answer is:
$\frac{13}{\sqrt{136}}$
Given circle,
$x^2+y^2-2 x+4 y-4=0$
and $x^2+y^2+4 x-6 y-3=0$
Equation of common chord of circle is $S_1-S_2=0$
$\therefore \quad\left(x^2+y^2-2 x+4 y-4\right)$
$-\left(x^2+y^2+4 x-6 y-3\right)=0$
$\begin{array}{ll}\Rightarrow & -6 x+10 y-1=0 \\ \Rightarrow & 6 x-10 y+1=0\end{array}$
Perpendicular distance from the point $(1,2)$ to the line $6 x-10 y+1=0$ is $\left|\frac{6-20+1}{\sqrt{6^2+10^2}}\right|$
$=\frac{13}{\sqrt{136}}$
$x^2+y^2-2 x+4 y-4=0$
and $x^2+y^2+4 x-6 y-3=0$
Equation of common chord of circle is $S_1-S_2=0$
$\therefore \quad\left(x^2+y^2-2 x+4 y-4\right)$
$-\left(x^2+y^2+4 x-6 y-3\right)=0$
$\begin{array}{ll}\Rightarrow & -6 x+10 y-1=0 \\ \Rightarrow & 6 x-10 y+1=0\end{array}$
Perpendicular distance from the point $(1,2)$ to the line $6 x-10 y+1=0$ is $\left|\frac{6-20+1}{\sqrt{6^2+10^2}}\right|$
$=\frac{13}{\sqrt{136}}$
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