Let circle C be the image of x² + y² – 2x + 4y – 4 = 0 in the line 2x – 3y + 5 = 0 and A be the point on C such that OA is parallel to x-axis and A lies on the right hand side of the centre O of C. If B(α,ß), with ß < 4, lies on C such that the length of the arc
AB is (1/6)^th of the perimeter of C, then $β - \sqrt{3} α$ is equal to

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$3 + \sqrt{3}$

$4 - \sqrt{3}$

answer is D.

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Detailed Solution

Centre (1, –2), r = 3
Reflection of (1, –2) about 2x – 3y + 5 = 0
$\begin{array}{l} \frac{x - 1}{2} = \frac{y + 2}{- 3} = \frac{- 2 (2 + 6 + 5)}{13} = - 2 \\ x = - 3, y = 4 \end{array}$
Equation of circle ‘C’
C : (x+3)² + (y– 4)² = 9
A.T.Q.
Question Image
$\begin{array}{l} ℓ (arcAB) = \frac{1}{6} \times 2 πr \\ rθ = \frac{1}{6} \times 2 πr \\ θ = \frac{π}{3} \\ (α + 6)^{2} + (β - 4)^{2} = 27 \\ \frac{(α + 3)^{2} \pm (β - 4)^{2} = 9}{(α + 6)^{2} - (α + 3)^{2} = 18} \\ \Rightarrow 6 α = - 9 \\ \Rightarrow α = \frac{- 3}{2}, β = (4 - \frac{3 \sqrt{3}}{2}) \\ ∴ β - \sqrt{3} α \\ (4 - \frac{3 \sqrt{3}}{2}) + \frac{3 \sqrt{3}}{2} \\ = 4 \end{array}$

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