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 \[x+2y=\,\pm \,2\sqrt{5}.\]You need to login to perform this action.
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