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(D) The circle touches the $x$-axis at $(a, 0)$ and lies below the $x$-axis,so its center is $(a, -r)$ and its radius is $r$,where $r > 0$.The equatio

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$(\alpha, \beta^2 - 4\gamma)$

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(D) The circle touches the $x$-axis at $(a, 0)$ and lies below the $x$-axis,so its center is $(a, -r)$ and its radius is $r$,where $r > 0$.The equation of the circle is $(x - a)^2 + (y + r)^2 = r^2$,which simplifies to $x^2 - 2ax + a^2 + y^2 + 2ry + r^2 = r^2$,or $x^2 + y^2 - 2ax + 2ry + a^2 = 0$.Comparing this with $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$,we get $\alpha = 2a$,$\beta = 2r$,and $\gamma = a^2$.The circle cuts an intercept of length $b$ on the $y$-axis. Setting $x = 0$ in the equation,we get $y^2 + \beta y + \gamma = 0$. The roots are $y_1, y_2$,and $|y_1 - y_2| = b$.Using the property of roots,$|y_1 - y_2| = \sqrt{\beta^2 - 4\gamma} = b$,so $b^2 = \beta^2 - 4\gamma$.Thus,the ordered pair $(2a, b^2)$ is equal to $(\alpha, \beta^2 - 4\gamma)$.

Let the equation of the circle,which touches the $x$-axis at the point $(a, 0), a > 0$ and cuts off an intercept of length $b$ on the $y$-axis,be $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below the $x$-axis,then the ordered pair $(2a, b^2)$ is equal to:

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