Q.
One of the diameters of the circle circumscribing the rectangle ABCD4y=x+7. If A and B are the points (−3,4) and (5,4) respectively, then find the area of rectangle.
Let O be the centre of circle and M be mid-point of AB.
Then, OM⊥AB⇒M(1,4)
Since, slope of AB=0
Equation of straight line MO is x=1 and equation of diameter is 4y=x+7. ⇒ Centre is (1,2).
Also, O is mid-point of BD ⇒(2α+5,2β+4)=(1,2)⇒α=−3,β=0 ∴AD=(−3+3)2+(4−0)2=4
and AB=64+0=8
Thus, area of rectangle =8×4=32 sq units