A) Find the equation of the circle. b) Find the equation of the normal to the circle at (3, 4).

The center $(h, k)$ is the midpoint of the diameter's endpoints $(-3, 5)$ and $(7, 9)$.

$$h = \frac{-3+7}{2} = \frac{4}{2} = 2$$ $$k = \frac{5+9}{2} = \frac{14}{2} = 7$$

The center of the circle is $(2, 7)$.

The radius $r$ is the distance from the center $(2, 7)$ to one of the endpoints, for example, $(7, 9)$.

$$r = \sqrt{(7-2)^2 + (9-7)^2}$$ $$r = \sqrt{(5)^2 + (2)^2}$$ $$r = \sqrt{25 + 4}$$ $$r = \sqrt{29}$$

The square of the radius is $r^2 = 29$.

The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the center $(2, 7)$ and $r^2 = 29$:

$$\boxed{(x-2)^2 + (y-7)^2 = 29}$$

Using the points $(2, 7)$ and $(3, 4)$:

$$m = \frac{4-7}{3-2} = \frac{-3}{1} = -3$$

Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(3, 4)$ and slope $m = -3$:

$$y - 4 = -3(x - 3)$$ $$y - 4 = -3x + 9$$ $$y = -3x + 9 + 4$$ $$y = -3x + 13$$ $$\boxed{y = -3x + 13}$$