Consider a parabola `y^(2)=4x` with vertex at A and focus at S. PQ is a chord of the parabola which is normal at point P. If the abscissa and the ordinate of the point P are equal, then the square of the length of the diameter of the circumcircle of triangle PSQ is

Answer

To solve the problem, we need to find the square of the length of the diameter of the circumcircle of triangle PSQ, where P is a point on the parabola \( y^2 = 4x \) such that the abscissa and ordinate of P are equal.

### Step-by-Step Solution:

1. **Identify the Coordinates of Point P:**
Since the abscissa and ordinate of point P are equal, we can set \( P = (a, a) \). We substitute this into the parabola's equation:
\[ a^2 = 4a \]
This simplifies to:
\[ a^2 - 4a = 0 \implies a(a - 4) = 0 \]
Thus, \( a = 0 \) or \( a = 4 \). Therefore, the points P can be \( (0, 0) \) or \( (4, 4) \).

2. **Find the Focus S and Vertex A:**
The vertex A of the parabola \( y^2 = 4x \) is at the origin \( (0, 0) \), and the focus S is at \( (1, 0) \).

3. **Determine the Normal at Point P:**
The slope of the tangent line at point \( P(a, a) \) can be found using the derivative of the parabola. The derivative \( \frac{dy}{dx} \) at point \( P \) is given by:
\[ y^2 = 4x \implies 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y} \]
At \( P(4, 4) \):
\[ \frac{dy}{dx} = \frac{2}{4} = \frac{1}{2} \]
The slope of the normal line is the negative reciprocal:
\[ m_{\text{normal}} = -2 \]
The equation of the normal line at point \( P(4, 4) \) is:
\[ y - 4 = -2(x - 4) \implies y = -2x + 12 \]

4. **Find Point Q:**
To find point Q, we need to find where the normal intersects the parabola again. Substitute \( y = -2x + 12 \) into the parabola's equation:
\[ (-2x + 12)^2 = 4x \]
Expanding and rearranging gives:
\[ 4x^2 - 48x + 144 = 4x \implies 4x^2 - 52x + 144 = 0 \]
Dividing through by 4:
\[ x^2 - 13x + 36 = 0 \]
Solving this quadratic using the quadratic formula:
\[ x = \frac{13 \pm \sqrt{(13)^2 - 4 \cdot 1 \cdot 36}}{2 \cdot 1} = \frac{13 \pm \sqrt{169 - 144}}{2} = \frac{13 \pm 5}{2} \]
This gives us:
\[ x = 9 \quad \text{or} \quad x = 4 \]
Since \( x = 4 \) corresponds to point P, we have \( x = 9 \) for point Q. Substituting back to find y:
\[ y = -2(9) + 12 = -6 \]
Thus, \( Q = (9, -6) \).

5. **Calculate the Diameter of the Circumcircle of Triangle PSQ:**
The circumradius \( R \) of triangle PSQ can be calculated using the formula:
\[ R = \frac{abc}{4K} \]
where \( a, b, c \) are the lengths of the sides of the triangle and \( K \) is the area of the triangle.

The lengths of the sides can be calculated as follows:
- Length PS:
\[ PS = \sqrt{(4 - 1)^2 + (4 - 0)^2} = \sqrt{9 + 16} = 5 \]
- Length SQ:
\[ SQ = \sqrt{(9 - 1)^2 + (-6 - 0)^2} = \sqrt{64 + 36} = 10 \]
- Length PQ:
\[ PQ = \sqrt{(9 - 4)^2 + (-6 - 4)^2} = \sqrt{25 + 100} = 5\sqrt{5} \]

The area \( K \) can be calculated using the determinant formula for the area of a triangle given vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
\[ K = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the points \( P(4, 4), S(1, 0), Q(9, -6) \):
\[ K = \frac{1}{2} \left| 4(0 + 6) + 1(-6 - 4) + 9(4 - 0) \right| = \frac{1}{2} \left| 24 - 10 + 36 \right| = \frac{1}{2} \left| 50 \right| = 25 \]

Now substituting into the circumradius formula:
\[ R = \frac{5 \cdot 10 \cdot 5\sqrt{5}}{4 \cdot 25} = \frac{250\sqrt{5}}{100} = \frac{5\sqrt{5}}{2} \]

The diameter \( D \) of the circumcircle is:
\[ D = 2R = 5\sqrt{5} \]
Therefore, the square of the diameter is:
\[ D^2 = (5\sqrt{5})^2 = 25 \cdot 5 = 125 \]

### Final Answer:
The square of the length of the diameter of the circumcircle of triangle PSQ is \( \boxed{125} \).