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If $m_{1}$ and $m_{2}$ are the slopes of tangents to the circle $x^{2}+y^{2}=4$ from the point $(3,2)$, then $m_{1}-m_{2}$ is equal to
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The correct answer is:
$\frac{12}{5}$
Equation of pair of tangents is $S S_{1}=T^{2}$
$\Rightarrow\left(x^{2}+y^{2}-4\right)(9+4-4)=(3 x+2 y-4)^{2}$
$\Rightarrow 5 y^{2}+16 y-12 x y+24 x-52=0$
$\therefore \quad m_{1}+m_{2}=\frac{-2 h}{b}=\frac{12}{5}$
and $\quad m_{1} m_{2}=0$
Now, $m_{1}-m_{2}=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}$
$=\sqrt{\left(\frac{12}{5}\right)^{2}-0}$
$=\frac{12}{5}$
$\Rightarrow\left(x^{2}+y^{2}-4\right)(9+4-4)=(3 x+2 y-4)^{2}$
$\Rightarrow 5 y^{2}+16 y-12 x y+24 x-52=0$
$\therefore \quad m_{1}+m_{2}=\frac{-2 h}{b}=\frac{12}{5}$
and $\quad m_{1} m_{2}=0$
Now, $m_{1}-m_{2}=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}$
$=\sqrt{\left(\frac{12}{5}\right)^{2}-0}$
$=\frac{12}{5}$
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