Question

A rectangle $ABCD$ is inscribed in a circle with a diameter lying along the line $3y=x+10$. If $A$ and $B$ are the points $(-6,7)$ and $(4,7)$ respectively then

• A. centre of the circle is $(-1,3)$
• B. centre of the circle $(-1,0)$
• C. area of the rectangle is $100sq.units$
• D. area of the rectangle is $80sq.units$

Answer 1

Answer: (A), (D)

Let $O(h,k)$ be the centre of the circle

Then $O(h,k)$ lies on the diameter $3y=x+10,$ so that $3k=h+10.$ Also, $OA=OB$ (radii of the same circle), so that

$(h+6{)}^2=(h-4{)}^2$. That is, $h=-1$ and therefore, $k=3$.

Hence radius of the circle is

$OA=\sqrt{(-1+6{)}^2+(3-7{)}^2}$

$=\sqrt{25+16}=\sqrt{41}$

so that $AC,$ the diameter of the circle, is $2\sqrt{41}$

.Now, $AB=\sqrt{(-6-4{)}^2+(7-7{)}^2}=10$

$\Rightarrow BC=\sqrt{164-100}=8$

Thus required area of the rectangle is $AB\times BC=10\times 8=80sq.units$