Question

A tangent PT is drawn to the circle x² + y² = 4 at the point P(√3, 1). If a straight line L which is perpendicular to PT is a tangent to the circle (x- 3)² + y² = 1, then a possible equation of L is

• A. x - √3y = 1
• B. x- √3y = 4
• C. x - √3y = -1
• D. x-√3y = 7

Answer 1

The Correct Option is A

To find the equation of a line \(L\) that is perpendicular to the tangent at point \(P(\sqrt{3}, 1)\) on the circle \(x^2 + y^2 = 4\) and tangent to the circle \((x-3)^2 + y^2 = 1\), we proceed as follows:

1. Verifying Point \(P\) on the First Circle:
The equation of the first circle is \(x^2 + y^2 = 4\). 
Check if \(P(\sqrt{3}, 1)\) lies on it:

\( (\sqrt{3})^2 + 1^2 = 3 + 1 = 4 \)
Since the equation holds, \(P\) is on the circle.

2. Equation of the Tangent at \(P\):
For a circle \(x^2 + y^2 = r^2\), the tangent at point \((x_1, y_1)\) is \(x x_1 + y y_1 = r^2\). 
Here, \((x_1, y_1) = (\sqrt{3}, 1)\), \(r^2 = 4\), so:

\( x \cdot \sqrt{3} + y \cdot 1 = 4 \)
\( \sqrt{3} x + y = 4 \)
The slope of this tangent is:

\( y = -\sqrt{3} x + 4 \implies \text{slope} = -\sqrt{3} \)

3. Slope of Line \(L\):
Since \(L\) is perpendicular to the tangent, its slope is the negative reciprocal of \(-\sqrt{3}\):

\( \text{slope of } L = \frac{1}{\sqrt{3}} \)

4. Equation of Line \(L\):
Let the equation of \(L\) be \(y = \frac{1}{\sqrt{3}} x + c\). 
Rewrite in standard form:

\( \sqrt{3} y = x + c \)
\( x - \sqrt{3} y + c = 0 \)
Or, adjusting the constant: \(x - \sqrt{3} y - c = 0\). For consistency, use:

\( x - \sqrt{3} y + \sqrt{3} c = 0 \)

5. Tangency to the Second Circle:
The second circle is \((x-3)^2 + y^2 = 1\), with center \((3, 0)\) and radius 1. 
The distance from the center \((3, 0)\) to line \(L: x - \sqrt{3} y + \sqrt{3} c = 0\) must equal the radius (1):

\( \text{Distance} = \frac{|3 \cdot 1 + (-\sqrt{3}) \cdot 0 + \sqrt{3} c|}{\sqrt{1^2 + (-\sqrt{3})^2}} = \frac{|3 + \sqrt{3} c|}{\sqrt{1 + 3}} = \frac{|3 + \sqrt{3} c|}{2} = 1 \)
\( |3 + \sqrt{3} c| = 2 \)
Solve:

\( 3 + \sqrt{3} c = \pm 2 \)
\( \sqrt{3} c = -3 \pm 2 \)
\( c = \frac{-3 \pm 2}{\sqrt{3}} \)
Case 1: \( c = \frac{-3 + 2}{\sqrt{3}} = -\frac{1}{\sqrt{3}} \)
Case 2: \( c = \frac{-3 - 2}{\sqrt{3}} = -\frac{5}{\sqrt{3}} \)

6. Equations of Possible Lines:
For \(c = -\frac{1}{\sqrt{3}}\), the line is:

\( y = \frac{1}{\sqrt{3}} x - \frac{1}{\sqrt{3}} \)
\( \sqrt{3} y = x - 1 \)
\( x - \sqrt{3} y = 1 \)
For \(c = -\frac{5}{\sqrt{3}}\), the line is:

\( y = \frac{1}{\sqrt{3}} x - \frac{5}{\sqrt{3}} \)
\( \sqrt{3} y = x - 5 \)
\( x - \sqrt{3} y = 5 \)

7. Selecting the Equation:
Both lines are tangent to the second circle and perpendicular to the tangent at \(P\). A possible equation, as specified, is:

\( x - \sqrt{3} y = 1 \)

Final Answer:
The equation of the line \(L\) is \(x - \sqrt{3} y = 1\).