Circles
Circle
nta_pyq_2025_jan
Grade 11
Question:
Let the equation of the circle, which touches x-axis at the point (a, 0), a > 0 and cuts off an intercept of length b on y-axis be x + y - \alphax + \betay + \gamma = 0. If the circle lies below x-axis, then the ordered pair (2a, b ) is equal to 2 2 2
(\gamma, \beta - 4\alpha) 2
(\alpha, \beta + 4\gamma) 2
(\gamma, \beta + 4\alpha) 2
(\alpha, \beta - 4\gamma) 2
Step-by-Step Solution
Key Concept: Apply the core result for circle equations and tangents and simplify using the given constraints.
(4) 2 By pytogorus r = a + 2 2 b 4 = P 2 2 2 4a +b r = \sqrt 4 Equation of circle is (x - \alpha) + (y - \beta) = r 2 2 2 2 2 2 2 2 x + y - 2ax - 2py + \alpha + p - r = 0 comparision x + y - \alphax + \betay + r = 0 2 2 2 -\alpha = -2a, \beta = -2p, r = a 2 2 2 \Rightarrow 2a = \alpha, 4a + b = 4p 2 2 2 \alpha + b = 4p 2 2 2 \alpha + b = \beta So, (2a, b ) = (\alpha, \beta - 4r) 2 2
Correct Answer: 4