1. A circle has the equation $(x+3)^{2}+(y-4)^{2}=d$. What is the value of $d$ such that the $x$-axis is a tangent to the circle?

Step 1 ：The equation of the circle is given as \((x+3)^{2}+(y-4)^{2}=d\). The center of the circle is (-3, 4).

Step 2 ：The x-axis is a tangent to the circle, which means that the center of the circle lies on the line parallel to x-axis at a distance equal to the radius of the circle.

Step 3 ：The distance from the center to the x-axis is equal to the y-coordinate of the center, which is 4. This distance is also the radius of the circle.

Step 4 ：The equation of the circle is \((x-h)^{2}+(y-k)^{2}=r^{2}\), where (h, k) is the center and r is the radius.

Step 5 ：So, the value of d is the square of the radius, which is \(4^{2}=16\).

Step 6 ：Final Answer: The value of d such that the x-axis is a tangent to the circle is \(\boxed{16}\).