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:
Step-by-Step Solution
Key Concept: Centre of circle: $(5,3)$, $r^2=25-30+... = 4$, so $r=2$. The side parallel to $y=x+3$ (slope 1) means sides are along $y=x+c$ and $x+y+d=0$. Distance from centre to each side $=r/\sqrt{2}=\sqrt{2}$. Find the 4 vertices then compute $\sum(x_i^2+y_i^2)$.
Vertices $(5,5),(3,3),(7,3),(5,1)$. $\sum(x_i^2+y_i^2)=50+18+58+26=152$.
Correct Answer: 2