A circle has center P(5,5) and radius 5. Find its equation in the form x2 + y2 + cx + dy + e = 0.

The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Given center $P(5,5)$, so $h=5$ and $k=5$.
Given radius $r=5$.

Here, $a=1, b=1, c=-10, d=-10, e=25$.

Let's assume, for demonstration purposes, that point Q has coordinates $(x_Q, y_Q)$. (These coordinates would be given in the missing diagram).
The center of the circle is $P(5,5)$.

The gradient $m_{PQ}$ is given by the formula:

This would yield a specific numerical value if $(x_Q, y_Q)$ were known.

The tangent at any point on a circle is perpendicular to the radius at that point.
If $m_{PQ}$ is the gradient of the radius PQ, then the gradient of the tangent, $m_{tangent}$, is the negative reciprocal of $m_{PQ}$:

Using the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with point $Q(x_Q, y_Q)$ and gradient $m_{tangent}$:

This equation would then be rearranged into the form $ax + by = c$.