Tangent to the circle at points A and B from any external point P is represented by pair of straight lines x² - 3y² - 2x + 1 = 0. If O is the center of the circle then ∠ AOB will be

Concept:

1. Tangent is perpendicular to the radius at the point of contact.

2. Angle between two lines: The angle θ between the lines having slope m1 and m2 is given by

\(\tan \theta = \;\left| {\frac{{{m_2} - {m_1}}}{{1 + \;{m_1}\;{m_2}}}} \right|\)

3. Sum of interior angles of a quadrilateral = 360° 

Calculation:

Given that, tangent to the circle at points A and B is represented by pair of straight lines PA & PB

x2 - 3y2 - 2x + 1 = 0 -----(1)

From equation (1)

x2 - 3y2 - 2x + 1 = 0

⇒ (x - 1)² = 3y²

⇒ x - 1 = ± √3y

Therefore, eqation of tangent PA and PB are

x + √3y - 1 = 0    ----(2)

x - √3y - 1 = 0    ----(3)

We know that, slope of line ax + by + c = 0 is -a/b. Therefore,

The slope of line (2) and (3) are -1/√3 and 1/√3 

Therefore, angle b/w tangents PA & PB

\(\theta \ =\ tan^{-1}[\frac{\frac{1}{\sqrt 3}\ -\ (\frac{-1}{\sqrt 3})}{1\ +\ (\frac{1}{\sqrt 3})(\frac{-1}{\sqrt 3})}]\)

\(⇒ \theta \ =\ tan^{-1}(\sqrt 3)\ =\ 60^∘\)

We know that,

∠ O + ∠ A + ∠ P + ∠ B = 360^∘ 

⇒ ∠ O = 360∘ - 180∘ - 60∘ 

⇒ ∠ O = 120∘

Hence, option 4 is correct.