Question

One of the diameters of the circle circumscribing the rectangle ABCD is $4yx+7$. If $A$ and $B$ are the points

Answer 1

Let co-ordinate of $O\cong (\alpha ,\beta )$

Thus $4\beta =\alpha +7$

$\beta =\frac{\alpha +7}{4}$

$\because O\cong (\alpha ,\frac{\alpha +7}{4})$

as $O$ is the centre of circle if is clear that

$OA=OB=OD=OC=r$ (radius of circle)

$OA=OB$

using Distance formula

$(\alpha +3{)}^2+{(\frac{\alpha +7}{4}-4)}^2=(\alpha +3{)}^2+{(\frac{\alpha -5}{4}-4)}^2$

$(\alpha +3{)}^2=(\alpha -5{)}^2$

$16\alpha =16$

$\alpha =1$

$\because O\cong (\alpha ,\frac{\alpha +7}{4})\cong (1,2)$

as $O$ is center it is mid-point of $D$ & $B$

Thus $\frac{h+5}{2}=1$ & $\frac{K+4}{2}=2$

$h=-3K=0$

$D\cong (-3,0)$

area of rectangle $=DA\times AB$

$=\sqrt{(-3)-(-3{)}^2+(0-4{)}^2}\sqrt{(-3-5{)}^2+(4-4{)}^2}$

$=4\times 8=32$ square units.