1. Find the circle’s center and radius

Given: $x^2+y^2-10x-6y+30=0$

Complete the squares: $x^2-10x+y^2-6y+30=0$ $(x-5)^2-25+(y-3)^2-9+30=0$ $(x-5)^2+(y-3)^2=4$

So the circle has center $C=(5,3)$ and radius $r=2.$

2. Use the fact that a square is inscribed in the circle

For a square inscribed in a circle of radius $r$, the diagonal of the square equals the diameter: $\text{diagonal}=2r=4.$

Hence side length $a=\frac{4}{\sqrt2}=2\sqrt2.$

The center of the square is the same as the center of the circle, i.e. $(5,3)$.

3. Determine the directions of the sides

One side is parallel to $y=x+3,$ which has slope $1$.

So one pair of sides has direction vector proportional to $(1,1)$, and the adjacent pair has slope $-1$, with direction vector proportional to $(1,-1)$.

Unit vectors along these directions are $\mathbf{u}=\frac{1}{\sqrt2}(1,1),\qquad \mathbf{v}=\frac{1}{\sqrt2}(1,-1).$

Since the side length is $2\sqrt2$, the half-side vectors from the center to the midpoint-offset are $\frac{a}{2}\mathbf{u}=\sqrt2\cdot \frac{1}{\sqrt2}(1,1)=(1,1),$ $\frac{a}{2}\mathbf{v}=\sqrt2\cdot \frac{1}{\sqrt2}(1,-1)=(1,-1).$

Thus the vertices are obtained from the center $(5,3)$ as $C\pm (1,1)\pm (1,-1).$

Compute them:

• $(5,3)+(1,1)+(1,-1)=(7,3)$
• $(5,3)+(1,1)-(1,-1)=(5,5)$
• $(5,3)-(1,1)+(1,-1)=(5,1)$
• $(5,3)-(1,1)-(1,-1)=(3,3)$

So the four vertices are