1. Given side of the rectangle

One side has endpoints $A(1,2), \quad B(3,6).$

So, $\overrightarrow{AB}=(2,4).$ Hence the length of this side is $AB=\sqrt{(3-1)^2+(6-2)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}.$

2. Rectangle inscribed in a circle

A rectangle inscribed in a circle has its diagonals as diameters of the circle. Also, opposite side $CD$ is parallel and equal to $AB$.

Let the other side vector be $\overrightarrow{AD}=(p,q)$. Since adjacent sides of a rectangle are perpendicular, $\overrightarrow{AB}\cdot\overrightarrow{AD}=0.$ So, (2,4)\cdot(p,q)=0 \implies 2p+4q=0 \implies p+2q=0. \tag{1}

3. Use the given diameter line

A diagonal of the rectangle is a diameter of the circumcircle. The given diameter lies on the line $2x-y+4=0.$

So one of the diagonals of the rectangle must lie on this line.

Check endpoints:

• For $A(1,2)$: $2(1)-2+4=4\ne 0$
• For $B(3,6)$: $2(3)-6+4=4\ne 0$

Thus side $AB$ is not on the line, so the diameter line must be along diagonal $AC$ or $BD$.

Take diagonal $AC$. Then $C=B+\overrightarrow{AD}=(3+p,6+q).$ So, $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=(2+p,4+q).$

Since diagonal $AC$ lies along the line $2x-y+4=0$, its direction vector must be parallel to that line.

The line $2x-y+4=0 \implies y=2x+4$ has direction vector $(1,2)$.

Hence, (2+p,4+q)=\lambda(1,2). \tag{2}

From (2), 4+q=2(2+p)=4+2p \implies q=2p. \tag{3}

Using (1): $p+2q=0.$ Substitute $q=2p$: $p+4p=0 \implies 5p=0 \implies p=0, q=0,$ which is impossible for a rectangle.

So $AC$ is not the diameter on the given line.

4. Take the other diagonal $BD$ on the given line

Now, $D=A+\overrightarrow{AD}=(1+p,2+q).$ Thus, $\overrightarrow{BD}=\overrightarrow{D}-\overrightarrow{B}=(p-2,q-4).$

Since $BD$ lies along the given diameter line, its direction vector is parallel to $(1,2)$: (p-2,q-4)=\lambda(1,2). \tag{4} So, q-4=2(p-2)=2p-4 \implies q=2p. \tag{5}

Again from perpendicularity, $p+2q=0.$ Substitute $q=2p$: $p+4p=0 \implies p=0, q=0,$ again impossible.

So instead of only using direction, we should use the fact that the diameter line passes through the center, i.e. the midpoint of a diagonal.

5. Center of rectangle lies on the given diameter

The center of the rectangle is the midpoint of diagonal $AC$ (or $BD$). Let $\overrightarrow{AD}=(p,q),$ with perpendicularity condition 2p+4q=0 \implies p=-2q. \tag{6}

Vertices are: $A(1,2),\; B(3,6),\; D(1+p,2+q),\; C(3+p,6+q).$

Center of rectangle: $M=\text{midpoint of }AC=\left(\frac{1+3+p}{2},\frac{2+6+q}{2}\right)=\left(2+\frac p2,4+\frac q2\right).$

Since the given line is a diameter of the circle, it must pass through the center $M$: $2\left(2+\frac p2\right)-\left(4+\frac q2\right)+4=0.$ Simplify: $4+p-4-\frac q2+4=0$ p-\frac q2+4=0. \tag{7}

Using (6), $p=-2q$: $-2q-\frac q2+4=0$ $-\frac{5q}{2}+4=0$ $q=\frac{8}{5}.$ Then $p=-2q=-\frac{16}{5}.$

So,