1. Write the general circle and identify its center/radius

   Given $x^2+y^2-\alpha x+\beta y+\gamma=0.$

   Comparing with $x^2+y^2+2gx+2fy+c=0,$ we get $2g=-\alpha,\qquad 2f=\beta,\qquad c=\gamma.$

   Hence the center is $(-g,-f)=\left(\frac{\alpha}{2},-\frac{\beta}{2}\right),$ and radius $r=\sqrt{g^2+f^2-c}=\sqrt{\frac{\alpha^2}{4}+\frac{\beta^2}{4}-\gamma}.$

2. Use the condition that the circle touches the $x$-axis at $(a,0)$ and lies below the $x$-axis

   If a circle touches the $x$-axis at $(a,0)$ and lies below the $x$-axis, then its center must be vertically below that point.

   So center is $(a,-r).$

   Comparing with the center from the equation, $\left(\frac{\alpha}{2},-\frac{\beta}{2}\right)=(a,-r).$

   Therefore, $\frac{\alpha}{2}=a \quad\Rightarrow\quad \alpha=2a.$

3. Find the intercept cut on the $y$-axis

   On the $y$-axis, $x=0$. So the circle meets the $y$-axis where $y^2+\beta y+\gamma=0.$

   If the roots are $y_1,y_2$, then the length of intercept on the $y$-axis is $b=|y_1-y_2|.$

   For a quadratic $y^2+\beta y+\gamma=0$, the square of the difference of roots is $(y_1-y_2)^2 = \beta^2-4\gamma.$

   Hence, $b^2=\beta^2-4\gamma.$

4. Form the ordered pair

   We have $2a=\alpha, \qquad b^2=\beta^2-4\gamma.$

   Therefore, $(2a,b^2)=\left(\alpha,\beta^2-4\gamma\right).$

5. Match with options

   This is Option B.

Comparison with stored correct answer: Stored answer is B, which matches our result.