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Jee Main 2025 — Medium JEE math MCQ

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 \( x^2 + y^2 - \alpha x + \beta y + \gamma = 0 \). If the circle lies below \( x \)-axis, then the ordered pair \((2a, b^2)\) is equal to
  1. A. \( (\gamma, \beta^2 - 4\alpha) \)
  2. B. \( (\alpha, \beta^2 + 4\gamma) \)
  3. C. \( (\gamma, \beta^2 + 4\alpha) \)
  4. D. \( (\alpha, \beta^2 - 4\gamma) \)

Solution

The correct option is **D**. (D. \( (\alpha, \beta^2 - 4\gamma) \))

MATH

medium • PYQ Reworded
Question
Read carefully, then pick the best option.
Let the equation of the circle, which touches x x -axis at the point (a,0) (a, 0) , a>0 a > 0 and cuts off an intercept of length b b on y y -axis be x2+y2−αx+βy+γ=0 x^2 + y^2 - \alpha x + \beta y + \gamma = 0 . If the circle lies below x x -axis, then the ordered pair (2a,b2)(2a, b^2) is equal to
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Jee Main 2025 — Medium JEE Mathematics MCQ | MyGoalPrep