Mathematics · 2025
JEE Main · 28 January 2025, Shift 1 · Q12
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-α x+β…
Let the equation of the circle, which touches x -axis at the point $\displaystyle (\mathrm{a}, 0), \mathrm{a}>0$ and cuts off an intercept of length $\displaystyle b$ on $\displaystyle y$-axis be $\displaystyle x^2+y^2-\alpha x+\beta y+\gamma=0$. If the circle lies below $\displaystyle x$-axis, then the ordered pair $\displaystyle \left(2 \mathrm{a}, \mathrm{b}^2\right)$ is equal to
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \left(\alpha, \beta^2-4 \gamma\right)$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.