Circles
Circle
nta_pyq_2025_jan
Grade 11
Question:
Let circle C be the image of x + y - 2x + 4y - 4 = 0 in the line 2x - 3y + 5 = 0 and A be the point on C 2 2 such that OA is parallel to x-axis and A lies on the right hand side of the centre O of C . If B(\alpha, \beta), with \beta < 4, lies on C such that the length of the are AB is (1/6) th of the perimeter of C , then \beta - \sqrt3\alpha is equal to
3 + \sqrt3
4
4 - \sqrt3
3
Step-by-Step Solution
Key Concept: Apply the core result for circle equations and tangents and simplify using the given constraints.
(2) Centre (1, -2), r = 3 Reflection of (1, -2) about 2x - 3y + 5 = 0 x - 1 y + 2 -2(2 + 6 + 5) = = = -2 2 -3 13 x = -3, y = 4 Equation of circle ' C ' 2 2 C : (x + 3) + (y - 4) = 9 A.T.Q. 1 ℓ(arcAB) = \times 2\pir 6 1 r\theta = \times 2\pir 6 \pi \theta = 3 2 2 (\alpha + 6) + (\beta - 4) = 27 2 2 (\alpha + 3) $\pm$ (\beta - 4) = 9 2 2 (\alpha + 6) - (\alpha + 3) = 18 \Rightarrow 6\alpha = -9 -3 3\sqrt 3 \Rightarrow \alpha = , \beta = (4 - ) 2 2 \therefore \beta - \sqrt3\alpha 3\sqrt 3 3\sqrt 3 (4 - ) + 2 2 = 4
Correct Answer: 2