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

Answer: (D)

152

Answer 2

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