Question

One of the diameter of the circle circumscribing the rectangle $ABCD$ is $4y=x+7$. If $A$ and $B$ are the points $(-3,4)$ and $(5,4)$. respectively, then the area of the rectangle is

• A. $21$
• B. $32$
• C. $35$
• D. None of these

Answer 1

The correct option is B $32$

Let the equation of the circle circumscribing the rectangle $ABCD$ be

$x^2+y^2+2gx+2fy+c=0$ ...(1)

Since $A(-3,4)$ and $B(5,4)$ lies on (1), we get

$25-6g+8f+c=0$ ...(2)

and, $41+10+8f+c=0$ ...(3)

Also, centre of (1) lies on $4y=x+7$

$\therefore -g+4f+7=0$ ...(4)

Subtracting (2) from (3), we get

$16+16g=0\Rightarrow g=-1$ ...(5)

Solving (4) and (5), we get $f=-2.$

Substituting the values of $g$ and $f$ in (3), we get

$41-10-16+c=0\Rightarrow c=-15.$

$\therefore$ The equation of the circle (1) becomes

$x^2+y^2-2x-4y-15=0.$

Radius of the circle is $=\sqrt{1+4+15}=2\sqrt{5}.$

Since the rectangle is inscribed in the circle, Its diagonal will be the diameter of the circle.

$\therefore$ The length of the diagonal of the rectangle $=2\left(2\sqrt{5}\right)=4\sqrt{5}.$

Also, the length of the rectangle

$=AB=\sqrt{{(5+3)}^2+{(4-4)}^2}=8.$

$\therefore$ Its breadth $=\sqrt{{\left(4\sqrt{5}\right)}^2-{\left(8\right)}^2}=\sqrt{80-64}=\sqrt{16}=4.$

hence, the area of rectangle $=8\times 4=32.$