JEE Math Practice 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 $y-a x i s$ be $x^2+y^2-\alpha x+\beta y+\gamma=0$. If the circle lies below $x-a x i s$, then the ordered pair $\left(2 a, b^2\right)$ is equal to
- A.$\left(\alpha, \beta^2+4 \gamma\right)$
- B.$\left(\alpha, \beta^2-4 \gamma\right)$
- C.$\left(\gamma, \beta^2-4 \alpha\right)$
- D.$\left(\gamma, \beta^2+4 \alpha\right)$
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