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

Expert Answers
justaguide eNotes educator| Certified Educator

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

The center is the mid point or [(18 + 4)/2 , (-13 - 3)/2]

=> (11 , -8)

The radius of the circle is half the distance between the points. This is: (1/2)*sqrt [(18 - 4)^2 + (-13 + 3)^2]

=> (1/2)*sqrt [ 14^2 + 10^2]

=> (1/2)*sqrt 296

The equation of the circle is (x - 11)^2 + (y + 8)^2 = 74

The area of the circle is pi*74

giorgiana1976 | Student

We'll put the equation of the circle in the standard form:

(x-h)^2 + (y-k)^2 = r^2

The center of the circle has the coordinates (h,k) and the radius is r.

Since the given points are located on the circle, substituted in the equation of the circle, they verify it.

(18-h)^2 + (-13-k)^2 = r^2 (1)

(4-h)^2 + (-3-k)^2 = r^2 (2)

We could also determine the value of the radius, since the given points are the endpoints of the diameter.

D^2 = (18-4)^2 + (-13+3)^2

D^2 = 14^2 + 10^2

D^2 = 196 + 100

D^2 = 296

D = 2sqrt74

r = D/2

r = sqrt 74

We'll susbtitute r in (1) and (2):

(18-h)^2 + (-13-k)^2 = 74 (3)

(4-h)^2 + (-3-k)^2 = 74 (4)

We'll put (3) = (4):

(18-h)^2 + (-13-k)^2 = (4-h)^2 + (-3-k)^2

We'll expand the squares:

324 - 36h + h^2 + 169 + 26k + k^2 = 16 + 8h + h^2 + 9 + 6k + k^2

We'll eliminate and combine like terms:

817 - 36h + 26k = 25 + 8h + 6k

We'll move all terms to the left side:

-44h + 20k + 792 = 0

h = (18+4)/2

h=11

k = (-13-3)/2

k = -8

The equation of the circle is:

(x-11)^2 + (y+8)^2 = 74