question_answer

The line \[2x-y+1=0\] is a tangent to the circle at the point (2, 5) and the centre of the circles lies on \[x-2y=4.\] The radius of the circle is

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 \[\Delta PAO,\] \[\tan \theta =\frac{r}{PA}\]

\[\Rightarrow \]   \[\frac{3}{4}=\frac{r}{4\sqrt{5}}\]

\[\Rightarrow \]   \[r=3\sqrt{5}\]