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: \[ \sum(x_i^2 + y_i^2) = 152 \]