MHT CET · Maths · Circle
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
- A \(8\)
- B \(9\)
- C \(10\)
- D \(11\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
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
\(
\begin{aligned}
& \Rightarrow\left|\frac{1-2(2)-m}{\sqrt{1+4}}\right| < \sqrt{5} \\
& \Rightarrow|\mathrm{m}+3| < 5 \\
& \Rightarrow-5 < \mathrm{m}+3 < 5 \\
& \Rightarrow-8 < \mathrm{m} < 2
\end{aligned}
\)
\(\therefore\) 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
\(
\begin{aligned}
& \Rightarrow\left|\frac{1-2(2)-m}{\sqrt{1+4}}\right| < \sqrt{5} \\
& \Rightarrow|\mathrm{m}+3| < 5 \\
& \Rightarrow-5 < \mathrm{m}+3 < 5 \\
& \Rightarrow-8 < \mathrm{m} < 2
\end{aligned}
\)
\(\therefore\) The number of possible values of \(\mathrm{m}=9\)
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