Area of rectangle in which circle is inscribed

Question

One of the diameters of the circle circumscribing the rectangle ABCD is $4y= x + 7$. If A & B are the points (–3, 4) & (5,4) respectively, then we have to find the area of the rectangle.

I tried it a lot

One of it side's length and equation is 8 ad y=4

Now what to do next

Answer 1

The centre of the circle is the point of intersection of the diameter $4y =x +7$ and the perpendicular bisector of $AB $ which comes out to be $x=1$. Hence, the center is at $(1,2) $. Also, $(1,2)$ is the midpoint of $BD$, thus $D (-3,0) $.

Thus area of rectangle $ABCD =2 \times$ area of $\triangle ABD =2\times \frac {1}{2} \times AB\times AD = 8\times 4 =32$. Hope it helps.

Answer 2

Area of ABCD = 2 × Area of triangle ABC

$$ \mathrm{Area}(\Delta ABC) = \frac 17 \int_{-3}^{5} (x + 7) dx - \int_{-3}^{5} 4 dx $$

Answer 3

The center of the circle is on the perpendicular bisector of $AB$: $x=1$. It intersects the given diameter in $O(1,2)$. Thus you only have to reflect $A$ and $B$ with respect to $O$, you get the remaining two vertices: $A'(5,0)$, $B'(-3,0)$.