To find the value of r², we need to first determine the coordinates of the endpoints of the latus rectum of the parabola.

The general equation of a parabola is given by y² = 4ax, where 'a' is the distance between the focus and the vertex of the parabola. In this case, since the equation is y² = 4x, we can see that a = 1.

The latus rectum of a parabola passes through the focus and is perpendicular to the axis of symmetry. The distance between the focus and the vertex is equal to a, so the coordinates of the focus are (a, 0), which in this case is (1, 0).

Therefore, the coordinates of the endpoints of the latus rectum can be found by adding or subtracting the value of a from the x-coordinate of the focus. This gives us the coordinates (1 ± a, 0), which in this case are (1 ± 1, 0) or (0, 0) and (2, 0).

Now let's find the equations of the normals drawn at these endpoints.

The slope of the normal at any point (x₁, y₁) on the parabola is given by -1/(dy/dx), where dy/dx is the derivative of the equation of the parabola.

Differentiating y² = 4x with respect to x, we get:
2y * (dy/dx) = 4
dy/dx = 4/(2y) = 2/y

So, at the endpoints (0, 0) and (2, 0), the slopes of the normals are 0 and 1, respectively.

The equation of a line with slope m passing through a point (x₁, y₁) is given by:
(y - y₁) = m(x - x₁)

For the normal at (0, 0), the slope is 0. Plugging in the values, we get:
y - 0 = 0(x - 0)
y = 0

For the normal at (2, 0), the slope is 1. Plugging in the values, we get:
y - 0 = 1(x - 2)
y = x - 2

Now let's find the points of intersection of these normals with the circle.

Substituting the equation of the normal at (0, 0) into the equation of the circle, we get:
(x - 3)² + (0 - 2)² = r²
x² - 6x + 9 + 4 = r²
x² - 6x + 13 = r²

Substituting the equation of the normal at (2, 0) into the equation of the circle, we get:
(2 - 3)² + (y - 2)² = r²
1 + (y - 2)² = r²
(y - 2)² = r² - 1

Now we have two equations, one from each normal, that involve r². We need to find a common value of r² that satisfies both equations.

Substituting the value of x from the first equation into the second equation, we get:
(6x - x² - 13)