question_answer

The straight line \[x+2y=1\]meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

A) \[\frac{\sqrt{5}}{2}\]                 

B)                                       \[2\sqrt{5}\]

C)               \[\frac{\sqrt{5}}{4}\]                                         

D)               \[4\sqrt{5}\]

Correct Answer: A

Solution :

The straight line \[x+2y=1\]meets the coordinate axes at (1, 0) and (0, 1/2).   Since \[\angle AOB={{90}^{o}}\] \[\therefore \]Equation of circle is \[(x-1)(x-0)+(y-0)\left( y-\frac{1}{2} \right)=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-x-\frac{y}{2}=0\] Now, equation of tangent at origin to the given circle is \[2x+y=0\] Now, \[{{l}_{1}}=\frac{\left| 0+\frac{1}{2} \right|}{\sqrt{4+1}}=\frac{1}{2\sqrt{5}}\] Similarly,\[{{l}_{2}}=\frac{2}{\sqrt{5}}\] \[\therefore \]\[{{l}_{1}}+{{l}_{2}}=\frac{1}{2\sqrt{5}}+\frac{2}{\sqrt{5}}=\frac{\sqrt{5}}{2}\]