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Let $\mathrm{S}$ be the circumcircle of the triangle formed by the line $x-2 y-4=0$ with the coordinate axes. If $\mathrm{P}(-2,-4)$ is a point in the plane of the circle $S$ and $Q$ is a point on $S$ such that the distance between $\mathrm{P}$ and $\mathrm{Q}$ is the least, then $\mathrm{PQ}=$
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Verified Answer
The correct answer is:
$5 \sqrt{5}$
Line $x-2 y-4=0$ intersect the axes at $A(4,0)$ and $B(0,-2)$
$$
\begin{aligned}
& \therefore \text { Centre of circle }=\left(\frac{4+0}{2}, \frac{-2+0}{2}\right)=(2,-1) \\
& \text { radius }=\sqrt{(2-0)^2+(0+1)^2}=\sqrt{5} \\
& C P=\sqrt{(2+2)^2+(-1+4)^2}=5
\end{aligned}
$$
Distance of $P Q$ is least when points $P, Q$ and $C$ are collinear.
$$
\therefore \quad P Q=C P-C Q=5-\sqrt{5}
$$
$$
\begin{aligned}
& \therefore \text { Centre of circle }=\left(\frac{4+0}{2}, \frac{-2+0}{2}\right)=(2,-1) \\
& \text { radius }=\sqrt{(2-0)^2+(0+1)^2}=\sqrt{5} \\
& C P=\sqrt{(2+2)^2+(-1+4)^2}=5
\end{aligned}
$$
Distance of $P Q$ is least when points $P, Q$ and $C$ are collinear.
$$
\therefore \quad P Q=C P-C Q=5-\sqrt{5}
$$
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