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

The equation of tangent PT to circle x² + y² = 4 at the point P(√3, 1) is √3x + y = 4.
Let, slope of line L be m.
m × (-√3) = -1 (Since L is perpendicular to PT)
m = 1/√3
Any line with slope 1/√3 will be of the form y = (1/√3)x + c
Given that this line is a tangent to the circle (x - 1)² + y² = 1
⇒ |1/√3 - 1 + c|/√((1/√3)² + 1²) = 1
⇒ |c + √3|/2 = 1
⇒ c + √3 = ±2
⇒ c = -√3 ± 2
Equation of line 'L' is y = (1/√3)x + (-√3 ± 2)
√3y = x ± 2√3
√3y = x + 2√3 or √3y = x - 2√3