Equation of Perpendicular Line Tangent to Circle

To find a possible equation of the straight line $L$, we proceed step-by-step:

1. Find the equation and slope of the tangent $PT$: The equation of the circle is: $x^2 + y^2 = 4$ The point $P(\sqrt{3}, 1)$ lies on this circle since $(\sqrt{3})^2 + 1^2 = 4$. The equation of the tangent $PT$ at $P(\sqrt{3}, 1)$ is given by $T = 0$: $x(\sqrt{3}) + y(1) = 4 \implies \sqrt{3}x + y - 4 = 0$ The slope of the tangent $PT$ is: $m_{PT} = -\sqrt{3}$

2. Find the slope and general equation of line $L$: The line $L$ is perpendicular to $PT$. Therefore, its slope $m_L$ is: $m_L = -\frac{1}{m_{PT}} = \frac{1}{\sqrt{3}}$ Thus, the equation of line $L$ can be written in the form: $x - \sqrt{3}y + k = 0$

3. Apply the condition of tangency to the second circle: The line $L$ is tangent to the circle: $(x - 3)^2 + y^2 = 1$ This circle has its center at $C(3, 0)$ and radius $r = 1$. The perpendicular distance from the center $(3, 0)$ to the line $L: x - \sqrt{3}y + k = 0$ must be equal to the radius $r = 1$: $\frac{|1(3) - \sqrt{3}(0) + k|}{\sqrt{1^2 + (-\sqrt{3})^2}} = 1$ $\frac{|3 + k|}{\sqrt{1 + 3}} = 1$ $\frac{|3 + k|}{2} = 1 \implies |3 + k| = 2$

   Solving this absolute value equation: $3 + k = 2 \implies k = -1$ $3 + k = -2 \implies k = -5$

4. Determine the possible equations of $L$:

   • For $k = -1$: $x - \sqrt{3}y - 1 = 0 \implies x - \sqrt{3}y = 1$
   • For $k = -5$: $x - \sqrt{3}y - 5 = 0 \implies x - \sqrt{3}y = 5$

Comparing with the given options, the correct equation is $x - \sqrt{3}y = 1$, which corresponds to Option (A).

Correct Answer: (A)