One of the diameters of the circle circumscribing the rectangle $ABCD$is $4y = x + 7$. If A and Bare the points (-3,4) and (5, 4) respectively, then the area of the rectangle, in square units, is

Let the equation of the circle be

$x^2+y^2+2gx+2fy+c=0$

This passes through (- 3, 4) and (5, 4).

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-6g+8f+c+25=0\\ \text{ and, }10g+8f+c+41=0\end{array}$

Subtracting (ii) from (i), we get g = -1. Since the centre (- g, - f) lies on $4 y = x + 7$

$\therefore -4f=-g+7\Rightarrow$

So, centre of the circle is at (1, 2)

Now, $AD = 2 GM$

=> AD = 2 (Length of the .l from G on AB whose eqn. is

$\begin{array}{l}\Rightarrow AD=2\times 2=4\\ \text{ Also, }AB=8\end{array}$

Hence, area of rectangle $ABCD =c 4 x 8 = 32$ sq. units.