A rectangle $$R$$ with end points of the one of its sides as $$(1, 2)$$ and $$(3, 6)$$ is inscribed in a circle. If the equation of a diameter of the circle is $$2x - y + 4 = 0$$, then the area of $$R$$ is ______

A rectangle $$R$$ has one side with endpoints $$(1, 2)$$ and $$(3, 6)$$, is inscribed in a circle, and the circle has diameter $$2x - y + 4 = 0$$. Find the area of $$R$$.

The center of the circle lies on the diameter $$2x - y + 4 = 0$$ and, since for a rectangle inscribed in a circle the center is the midpoint of the diagonals and is equidistant from all four vertices, we proceed by finding key points.

The midpoint of $$(1, 2)$$ and $$(3, 6)$$ is $$M = (2, 4)$$.

Because the perpendicular bisector of any chord passes through the center, we compute the slope of the side from $$(1, 2)$$ to $$(3, 6)$$ as $$m = \frac{6-2}{3-1} = 2$$. Hence the perpendicular bisector has slope $$-\frac{1}{2}$$ and passes through $$(2, 4)$$. Its equation is $$y - 4 = -\frac{1}{2}(x - 2)$$, which simplifies to $$2y - 8 = -x + 2$$ and therefore $$x + 2y = 10$$.

Substituting $$2x - y + 4 = 0 \Rightarrow y = 2x + 4$$ into $$x + 2y = 10$$ gives $$x + 2(2x + 4) = 10$$, so $$x + 4x + 8 = 10$$ and thus $$5x = 2$$. Therefore, $$x = \frac{2}{5}, \quad y = 2 \cdot \frac{2}{5} + 4 = \frac{24}{5}$$ and the center is $$\left(\frac{2}{5}, \frac{24}{5}\right)$$.

It follows that the radius satisfies $$r^2 = \left(1 - \frac{2}{5}\right)^2 + \left(2 - \frac{24}{5}\right)^2 = \left(\frac{3}{5}\right)^2 + \left(-\frac{14}{5}\right)^2 = \frac{9}{25} + \frac{196}{25} = \frac{205}{25} = \frac{41}{5}$$.

The length of the given side is $$|AB| = \sqrt{(3-1)^2 + (6-2)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$$.

Since the diagonal of the rectangle equals the diameter, we have diameter = $$2r$$, so diagonal$$^2 = 4r^2 = \frac{4 \times 41}{5} = \frac{164}{5}$$. For a rectangle: diagonal$$^2$$ = side$$_1^2$$ + side$$_2^2$$, hence $$\frac{164}{5} = 20 + \text{side}_2^2$$, which gives $$\text{side}_2^2 = \frac{164}{5} - 20 = \frac{164 - 100}{5} = \frac{64}{5}$$ and $$\text{side}_2 = \frac{8}{\sqrt{5}}$$.

Therefore, the area of the rectangle is $$2\sqrt{5} \times \frac{8}{\sqrt{5}} = 16$$.

The answer is $$\boxed{16}$$.