A) \[3\sqrt{5}\]
B) \[5\sqrt{3}\]
C) \[2\sqrt{5}\]
D) \[5\sqrt{2}\]
Correct Answer: A
Solution :
Given, \[2x-y+1=0\]is tangent of circle at (2, 5). |
Centre of circle lies on \[x-2y=4\] |
Solving \[2x-y+1=0\]and \[x-2y=4,\] we get intersecting point \[P\,(-\,2,-\,3)\] |
\[PA=\sqrt{{{(2+2)}^{2}}+{{(5+3)}^{2}}}\] \[=\sqrt{16+64}=\sqrt{80}=4\sqrt{5}\] |
Angle between lines are \[\tan \theta =\frac{2-\frac{1}{2}}{1+2\left( \frac{1}{2} \right)}=\frac{3}{4}\] |
In \[\Delta PAO,\] \[\tan \theta =\frac{r}{PA}\] |
\[\Rightarrow \] \[\frac{3}{4}=\frac{r}{4\sqrt{5}}\] |
\[\Rightarrow \] \[r=3\sqrt{5}\] |
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