If the normals of the parabola `y^(2)=4x` drawn at the end points of its latus rectum are tangents to the circle `(x-3)^(2)+(y+2)^(2)=r^(2)`, then the value of `r^(4)` is equal to

Answer

To solve the problem, we need to follow these steps:

### Step 1: Identify the endpoints of the latus rectum of the parabola
The parabola given is \( y^2 = 4x \). The latus rectum of this parabola is vertical and passes through the focus, which is at \( (1, 0) \). The endpoints of the latus rectum can be found at \( (1, 2) \) and \( (1, -2) \).

### Step 2: Find the slopes of the normals at the endpoints
The slope of the tangent to the parabola at any point \( (x_0, y_0) \) is given by the derivative. For the parabola \( y^2 = 4x \), we differentiate:
\[ \frac{dy}{dx} = \frac{2y}{4} = \frac{y}{2} \]
At the point \( (1, 2) \):
\[ \text{slope of tangent} = \frac{2}{2} = 1 \quad \Rightarrow \quad \text{slope of normal} = -1 \]
At the point \( (1, -2) \):
\[ \text{slope of tangent} = \frac{-2}{2} = -1 \quad \Rightarrow \quad \text{slope of normal} = 1 \]

### Step 3: Write the equations of the normals
Using the point-slope form of the line, the equations of the normals at the endpoints are:
1. For point \( (1, 2) \):
\[ y - 2 = -1(x - 1) \quad \Rightarrow \quad y = -x + 3 \]
2. For point \( (1, -2) \):
\[ y + 2 = 1(x - 1) \quad \Rightarrow \quad y = x - 3 \]

### Step 4: Find the center and radius of the circle
The equation of the circle given is:
\[ (x - 3)^2 + (y + 2)^2 = r^2 \]
The center of the circle is \( (3, -2) \).

### Step 5: Find the distance from the center of the circle to the lines
To find the radius \( r \), we need to calculate the perpendicular distance from the center \( (3, -2) \) to each of the normals.

1. For the line \( y = -x + 3 \) (or \( x + y - 3 = 0 \)):
\[ \text{Distance} = \frac{|3 + (-2) - 3|}{\sqrt{1^2 + 1^2}} = \frac{|3 - 2 - 3|}{\sqrt{2}} = \frac{| - 2 |}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

2. For the line \( y = x - 3 \) (or \( -x + y + 3 = 0 \)):
\[ \text{Distance} = \frac{|-3 + (-2) + 3|}{\sqrt{(-1)^2 + 1^2}} = \frac{| - 2 |}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

Since both distances equal \( \sqrt{2} \), we have \( r = \sqrt{2} \).

### Step 6: Calculate \( r^4 \)
Now, we need to find \( r^4 \):
\[ r^2 = 2 \quad \Rightarrow \quad r^4 = (r^2)^2 = 2^2 = 4 \]

### Final Answer
Thus, the value of \( r^4 \) is \( \boxed{4} \).