A) \[3x+4y=\pm 2\sqrt{5}\]
B) \[6x+8y=\pm \sqrt{5}\]
C) \[3x+4y=\pm \sqrt{5}\]
D) None of the above
Correct Answer: C
Solution :
Given equation of circle is \[5{{x}^{2}}+5{{y}^{2}}=1\] or \[{{x}^{2}}+{{y}^{2}}=\frac{1}{5}\] centre of the circle is \[(0,\,0)\]. Equation of tangent which are parallel to \[3x+4y-1=0\] is \[3x+4y+\lambda =0\] ??(i) We know that perpendicular distance from centre \[(0,\,0)\] to \[3x+4y+\lambda =0\] should be equal to radius. \[\therefore \] \[\frac{3\times 0+4\times 0+\lambda }{\sqrt{{{(3)}^{2}}+{{(4)}^{2}}}}=\pm \frac{1}{\sqrt{5}}\] \[\Rightarrow \] \[\lambda =\pm \frac{5}{\sqrt{5}}=\pm \sqrt{5}\] On putting the value of \[\lambda \] in Eq. (i), we get \[3x+4y\pm \sqrt{5}=0\] or \[3x+4y=\pm \sqrt{5}\]
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