Question

A square is inscribed in the circle \( x^2 + y^2 - 10x - 6y + 30 = 0 \). One side of this square is parallel to \( y = x + 3 \). If \( (x_i, y_i) \) are the vertices of the square, then \( \sum (x_i^2 + y_i^2) \) is equal to:

• A. 148
• B. 156
• C. 160
• D. 152

Answer 1

The Correct Option is D

Rewrite the Equation of the Circle: 
The given equation of the circle is: \[ x^2 + y^2 - 10x - 6y + 30 = 0 \]
To find the center and radius, complete the square for \(x\) and \(y\): \[ (x^2 - 10x) + (y^2 - 6y) = -30 \] 
Completing the square:
\[ (x - 5)^2 - 25 + (y - 3)^2 - 9 = -30 \]
\[ (x - 5)^2 + (y - 3)^2 = 4 \] 
So, the circle has center \((5, 3)\) and radius \(2\). 

Properties of the Inscribed Square: 
Since the square is inscribed in the circle, its diagonal is equal to the diameter of the circle. 
The diameter of the circle is \(2 \times 2 = 4\), so the diagonal of the square is \(4\). 

Calculate the Side Length of the Square: 
For a square, if the diagonal length is \(d\), then the side length \(s\) is given by:
\[ s = \frac{d}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \] 
 

Determine the Vertices of the Square: 
Since one side of the square is parallel to the line \(y = x + 3\), the square is oriented at a 45-degree angle. The center of the square is the same as the center of the circle, \((5, 3)\). 
 

Using the center \((5, 3)\) and the side length \(2\sqrt{2}\), the vertices of the square can be determined as:
\[ \left(5 + \frac{2}{\sqrt{2}}, 3 + \frac{2}{\sqrt{2}}\right), \quad \left(5 - \frac{2}{\sqrt{2}}, 3 + \frac{2}{\sqrt{2}}\right), \]
\[ \left(5 + \frac{2}{\sqrt{2}}, 3 - \frac{2}{\sqrt{2}}\right), \quad \left(5 - \frac{2}{\sqrt{2}}, 3 - \frac{2}{\sqrt{2}}\right) \] 
 

Calculate \(\sum(x_i^2 + y_i^2)\): 
Each vertex \((x_i, y_i)\) of the square satisfies:
\[ x_i^2 + y_i^2 = \left(5 \pm \frac{2}{\sqrt{2}}\right)^2 + \left(3 \pm \frac{2}{\sqrt{2}}\right)^2 \]
Simplifying for each vertex and summing,
we get: \[ \sum(x_i^2 + y_i^2) = 152 \]

Answer 2

Step 1: Write the given circle in standard form.
The given circle is:
\[ x^2 + y^2 - 10x - 6y + 30 = 0 \] Complete the squares for \(x\) and \(y\):
\[ (x^2 - 10x) + (y^2 - 6y) = -30 \] \[ (x - 5)^2 - 25 + (y - 3)^2 - 9 = -30 \] \[ (x - 5)^2 + (y - 3)^2 = 4 \] Thus, the center of the circle is \(C(5, 3)\) and its radius \(r = 2\).

Step 2: Relation between circle and square.
If a square is inscribed in a circle, the diagonal of the square equals the diameter of the circle.
Therefore, the diagonal of the square is: \[ d = 2r = 4 \] Let the side of the square be \(a\). Then, from the relation \(d = a\sqrt{2}\), we get: \[ a\sqrt{2} = 4 \quad \Rightarrow \quad a = 2\sqrt{2}. \]

Step 3: Orientation of the square.
We are told that one side of the square is parallel to the line \(y = x + 3\).
The slope of this line is \(1\), meaning the sides of the square make a \(45^\circ\) angle with the x-axis.

Hence, the diagonals of the square will be along lines with slopes \(1\) and \(-1\). Since the circle’s center is also the square’s center, the equations of the diagonals are:
\[ y - 3 = (1)(x - 5) \quad \Rightarrow \quad y = x - 2, \] and \[ y - 3 = (-1)(x - 5) \quad \Rightarrow \quad y = -x + 8. \]

Step 4: Find the coordinates of the vertices of the square.
The diagonals are perpendicular bisectors, each passing through the center \( (5, 3) \).
Half the diagonal = \( \frac{d}{2} = 2 \).
So the endpoints of each diagonal are 2 units from the center along each diagonal.

For the diagonal along \( y = x - 2 \):
Direction vector = \( (1, 1) \). Unit vector = \( \frac{1}{\sqrt{2}}(1, 1) \).
The vertices on this diagonal are: \[ (5, 3) \pm 2 \times \frac{1}{\sqrt{2}}(1, 1) = (5, 3) \pm (\sqrt{2}, \sqrt{2}) \] Hence, the coordinates are \( (5 + \sqrt{2}, 3 + \sqrt{2}) \) and \( (5 - \sqrt{2}, 3 - \sqrt{2}) \).

For the other diagonal \( y = -x + 8 \):
Direction vector = \( (1, -1) \). Unit vector = \( \frac{1}{\sqrt{2}}(1, -1) \).
The vertices on this diagonal are: \[ (5, 3) \pm 2 \times \frac{1}{\sqrt{2}}(1, -1) = (5, 3) \pm (\sqrt{2}, -\sqrt{2}) \] Hence, the coordinates are \( (5 + \sqrt{2}, 3 - \sqrt{2}) \) and \( (5 - \sqrt{2}, 3 + \sqrt{2}) \).

Step 5: Compute \( \sum (x_i^2 + y_i^2) \).
We have the four vertices:
\[ (5 + \sqrt{2}, 3 + \sqrt{2}), \quad (5 - \sqrt{2}, 3 - \sqrt{2}), \quad (5 + \sqrt{2}, 3 - \sqrt{2}), \quad (5 - \sqrt{2}, 3 + \sqrt{2}) \] Compute \( x_i^2 + y_i^2 \) for each vertex:

For \( (5 + \sqrt{2}, 3 + \sqrt{2}) \): \[ x^2 + y^2 = (5 + \sqrt{2})^2 + (3 + \sqrt{2})^2 = (25 + 10\sqrt{2} + 2) + (9 + 6\sqrt{2} + 2) = 38 + 16\sqrt{2} \] For \( (5 - \sqrt{2}, 3 - \sqrt{2}) \): \[ x^2 + y^2 = (25 - 10\sqrt{2} + 2) + (9 - 6\sqrt{2} + 2) = 38 - 16\sqrt{2} \] For \( (5 + \sqrt{2}, 3 - \sqrt{2}) \): \[ x^2 + y^2 = (25 + 10\sqrt{2} + 2) + (9 - 6\sqrt{2} + 2) = 38 + 4\sqrt{2} \] For \( (5 - \sqrt{2}, 3 + \sqrt{2}) \): \[ x^2 + y^2 = (25 - 10\sqrt{2} + 2) + (9 + 6\sqrt{2} + 2) = 38 - 4\sqrt{2} \] Sum all four: \[ (38 + 16\sqrt{2}) + (38 - 16\sqrt{2}) + (38 + 4\sqrt{2}) + (38 - 4\sqrt{2}) = 38 \times 4 = 152 \]

Final Answer:
\[ \boxed{152} \]