In a rectangle ABCD, the coordinates of A and B are (1,2) and (3,6) respectively The diameter of the circumscribing circle of ABCD lies in the line 2x − y + 4 = 0. If the bisects two parallel sides of the rectangle, then the area of the rectangle is _______ squared units.

Given:

A(1, 2) and B(3, 6)

By using the distance formula:
$\begin{array}{l}\mathrm{AB}=\sqrt{{(3-1)}^2+{(6-2)}^2}\\ =\sqrt{2^2+4^2}\\ =\sqrt{4+16}\\ =\sqrt{20}\end{array}$ 

$=2\sqrt{5}$
We know, that the slope of a line is $\frac{-\mathrm{a}}{\mathrm{b}}$ in the equation $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ 
Therefore, the slope of diameter of given equation $2\mathrm{x}-\mathrm{y}+4=0$  is,$-\left(\frac{2}{-1}\right)=2$ 
Now, the slope of line joining $\mathrm{A}(1,2)$  and $\mathrm{B}(3,6)$  is 

 $\frac{6-2}{3-1}$

$=\frac{4}{2}$

$=2$ 
On drawing the figure:

To find BE using the point to line distance formula,

$\mathrm{BE}=\frac{\left|{\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{C}\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$
$\mathrm{BE}=\frac{2\left(3\right)-6+4}{\sqrt{2^2+{(-1)}^2}}$

$\Rightarrow \mathrm{BE} =\frac{4}{\sqrt{5}}$ 
As the diameter divides the side of the rectangle ,

Therefore, length of BC =
$2\mathrm{BE}=2\times \frac{4}{\sqrt{5}}=\frac{8}{\sqrt{5}}$ 
Area of rectangle:
$\mathrm{Area}=\mathrm{AB}\times \mathrm{BC}$

$=2\sqrt{5}\times \frac{8}{\sqrt{5}}$

$=16$ squared units.