Find the equation and the area of the circle if the ends of the diameter (18,-13) and (4,-3).

To find the equation and the area of the circle if the
ends of the diameter (18,-13) and (4,-3).

The midpoint of
the diameter is the centre of the circle.

The ends of the
diameter are A(18,-13 and B(4,-3).

Therefore the coordinate
of the centre C(x,y) is given by:

C(x,y) = ((Ax+By)/2,
(Ay+By)/2))= ({18+4)/2,(-13-3)/2))).

The  radius of the
circle = semidiameter .

Diameter = distance between A and B
=  sqrt{(Bx-Ax)^2+(By-Ay)^2} =  sqrt{((4-18)^2 +(-3-13)^2} = sqrt(196+100) =
sqrt296.

Therefore radius r of the given cicle = diameter/2
= sqrt296/2 = sqrt74.

r = sqrt
74.

C(x,y)=(11,-8), r =
sqrt74.

The equation of the circle with centre (h,k) and
radius r is (x-h)^2+(y-k)^2 = r^2. Substitute the value for (h,k) = C(11,-8) and r^2= 74
and we get:

(x-11)^2+(y-(-8))^2= 74 is the equation of the
circle with A(18,-13) and B(4,-3) as diameter. We convert this into another standard
form by expanding and rearrging the equation,(x-11)^2+(y-(-8))^2=
74.

x^2-22x+121+y^2+16x+64 =
74.

x^2+y^2-22x+16x-10 = 0 is the standard form of the
equation of the same circle .