Circle 3 Question 13
13. A possible equation of $L$ is
(a) $x-\sqrt{3} y=1$
(b) $x+\sqrt{3} y=1$
(c) $x-\sqrt{3} y=-1$
(d) $x+\sqrt{3} y=5$
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Answer:
Correct Answer: 13. (a)
Solution:
- Here, tangent to $x^{2}+y^{2}=4$ at $(\sqrt{3}, 1)$ is $\sqrt{3} x+y=4$…(i)
As, $L$ is perpendicular to $\sqrt{3} x+y=4$
$\Rightarrow x-\sqrt{3} y=\lambda$ which is tangent to
$$ \begin{array}{rlrl} & & (x-3)^{2}+y^{2} & =1 \\ \Rightarrow & & \frac{|3-0-\lambda|}{\sqrt{1+3}} & =1 \\ \Rightarrow & & |3-\lambda| & =2 \\ \Rightarrow & & 3-\lambda & =2,-2 \\ & \therefore & \lambda & =1,5 \\ \Rightarrow & L: x-\sqrt{3} y & =1, x-\sqrt{3} y=5 \end{array} $$
