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(B) Let the center of the circle be $O(h, k)$. Since the center lies on the line $3y = x + 10$,we have $3k = h + 10$,or $h = 3k - 10$. Since $ABCD$ is

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$80$

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(B) Let the center of the circle be $O(h, k)$. Since the center lies on the line $3y = x + 10$,we have $3k = h + 10$,or $h = 3k - 10$. Since $ABCD$ is a rectangle inscribed in a circle,its center $O$ is the midpoint of the diagonal $AC$ and $BD$. Also,the perpendicular bisector of any chord passes through the center. The points $A(-6, 7)$ and $B(4, 7)$ lie on the circle. The line segment $AB$ is horizontal because the $y$-coordinates are the same. The perpendicular bisector of $AB$ is the vertical line passing through the midpoint of $AB$. The midpoint of $AB$ is $(\frac{-6+4}{2}, \frac{7+7}{2}) = (-1, 7)$. Thus,the perpendicular bisector of $AB$ is $x = -1$. Since the center $O(h, k)$ lies on this perpendicular bisector,we have $h = -1$. Substituting $h = -1$ into $h = 3k - 10$,we get $-1 = 3k - 10$,which implies $3k = 9$,so $k = 3$. The center of the circle is $O(-1, 3)$. The length of side $AB$ is $|4 - (-6)| = 10$. The distance from the center $O(-1, 3)$ to the line $AB$ $(y=7)$ is $|7 - 3| = 4$. In a rectangle inscribed in a circle,the distance from the center to the side $AB$ is half the length of the other side $AD$. So,$\frac{AD}{2} = 4$,which means $AD = 8$. The area of the rectangle $ABCD = AB 	imes AD = 10 	imes 8 = 80$.

$A$ rectangle $ABCD$ is inscribed in a circle whose center lies on the line $3y = x + 10$. If $A$ and $B$ are the points $(-6, 7)$ and $(4, 7)$ respectively,find the area of the rectangle.

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