question_answer

The equation to the tangents to the circle \[{{x}^{2}}+{{y}^{2}}=4\], which are parallel to \[x+2y+3=0\], are [MP PET 2003]

A)            \[x-2y=2\]                                

B)            \[x+2y=\pm \,2\sqrt{3}\]

C)            \[x+2y=\pm \,2\sqrt{5}\]       

D)            \[x-2y=\pm \,2\sqrt{5}\]

Correct Answer: C

Solution :

           Trick: Only (b) and (c) lines are parallel to \[x+2y+3=0\]                         Also the line is a tangent to \[{{x}^{2}}+{{y}^{2}}=4\]                    Its distance from (0, 0) should be 2.                    Therefore c is the answer.               Alternative method:                    Centre of \[{{x}^{2}}+{{y}^{2}}=4\] is (0, 0).                    Tangents which are parallel to \[x+2y+3=0\] is \[x+2y+\lambda =0\]                                     .....(i)                    Perpendicular distance from (0,0) to \[x+2y+\lambda =0\] should be equal to radius of circle, (Clearly radius = 2).                    \[\therefore \] \[\frac{0+2\times \,0+\lambda }{\sqrt{{{1}^{2}}+{{2}^{2}}}}=\,\pm \,2\] \[\Rightarrow \] \[\lambda \,=\pm \,2\sqrt{5}\]                    Put the value of \[\lambda \,\] in (i),                    tangents of circle are \[x+2y=\,\pm \,2\sqrt{5}.\]