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If the line $x-2 y=\mathrm{m}(\mathrm{m} \in \mathrm{Z})$ intersects the circle $x^2+y^2=2 x+4 y$ at two distinct points, then the number of possible values of $\mathrm{m}$ are
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Verified Answer
The correct answer is:
$9$
Centre of circle is $(1,2)$ and radius $=\sqrt{1+4-0}=\sqrt{5}$
Since the line intersects the circle at two points, length of perpendicular from the centre $ < $ radius
$$
\begin{aligned}
& \Rightarrow\left|\frac{1-2(2)-m}{\sqrt{1+4}}\right| < \sqrt{5} \\
& \Rightarrow|\mathrm{m}+3| < 5 \\
& \Rightarrow-5 < \mathrm{m}+3 < 5 \\
& \Rightarrow-8 < \mathrm{m} < 2
\end{aligned}
$$
$\therefore \quad$ The number of possible values of $\mathrm{m}=9$
Since the line intersects the circle at two points, length of perpendicular from the centre $ < $ radius
$$
\begin{aligned}
& \Rightarrow\left|\frac{1-2(2)-m}{\sqrt{1+4}}\right| < \sqrt{5} \\
& \Rightarrow|\mathrm{m}+3| < 5 \\
& \Rightarrow-5 < \mathrm{m}+3 < 5 \\
& \Rightarrow-8 < \mathrm{m} < 2
\end{aligned}
$$
$\therefore \quad$ The number of possible values of $\mathrm{m}=9$
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