a square is inscribed in the circle `x^2 + y^2- 10x - 6y + 30 = 0.` One side of the square is parallel to `y = x + 3,` then one vertex of the square is :

Answer

To solve the problem of finding one vertex of the square inscribed in the circle given by the equation \( x^2 + y^2 - 10x - 6y + 30 = 0 \), with one side of the square parallel to the line \( y = x + 3 \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 10x - 6y + 30 = 0 \]
We can complete the square for both \(x\) and \(y\).

**Completing the square for \(x\):**
\[ x^2 - 10x \rightarrow (x - 5)^2 - 25 \]

**Completing the square for \(y\):**
\[ y^2 - 6y \rightarrow (y - 3)^2 - 9 \]

Substituting back into the equation:
\[ (x - 5)^2 - 25 + (y - 3)^2 - 9 + 30 = 0 \]
This simplifies to:
\[ (x - 5)^2 + (y - 3)^2 - 4 = 0 \]
Thus, we have:
\[ (x - 5)^2 + (y - 3)^2 = 4 \]
This represents a circle with center \( (5, 3) \) and radius \( r = 2 \).

### Step 2: Determine the Orientation of the Square
Since one side of the square is parallel to the line \( y = x + 3 \), the slope of this line is \( 1 \). Therefore, the sides of the square will also have a slope of \( 1 \) or \( -1 \).

### Step 3: Find the Coordinates of the Vertices
Let’s denote the vertices of the square as \( P, Q, R, S \). Since the square is inscribed in the circle, the center of the square will also be the center of the circle, which is \( (5, 3) \).

The distance from the center to any vertex (which is the radius of the circle) is \( 2 \).

### Step 4: Calculate the Coordinates
To find the coordinates of the vertices, we can use the fact that the distance from the center to a vertex is \( 2 \) and the angles corresponding to the slopes \( 1 \) and \( -1 \).

Using the angle of \( 45^\circ \) (since the slope is \( 1 \)), we can find the coordinates of one vertex \( P \) by moving \( 2 \) units at \( 45^\circ \):
\[ P = \left( 5 + 2 \cdot \cos(45^\circ), 3 + 2 \cdot \sin(45^\circ) \right) \]
\[ P = \left( 5 + 2 \cdot \frac{1}{\sqrt{2}}, 3 + 2 \cdot \frac{1}{\sqrt{2}} \right) \]
\[ P = \left( 5 + \sqrt{2}, 3 + \sqrt{2} \right) \]

### Step 5: Identify One Vertex
Thus, one vertex of the square is:
\[ \boxed{(5 + \sqrt{2}, 3 + \sqrt{2})} \]