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

hala718 | High School Teacher | (Level 1) Educator Emeritus

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Given endpoints of the diameter in a circle are ( 18. -13) and ( 4, -3).

We need to find the equation of the circle.

We will write into the circle standard equation.

( y-a)^2 + ( x-b)^2 = r^2 where ( a, b) is the center and r is the radius.

Given the end points of the diameter, then we know that the midpoint of the diameter is the center of the circle.

==> xm = (xA+xB) /2 = ( 18+4)/2 = 22/2 = 11

==> ym = (yA+yB)/2 = ( -13-3) /2 = -16/2 = -8

Then the center is m(11,-8).

==> Now we will calculate the length of the diameter.

D =  sqrt( 18-4)^2 + ( -13+3)^2

= sqrt( 14^2 + 10^2)

= sqrt(196+100) = sqrt(296) =  17.2

==> r = d/2 = 17.2/2 = 8.6

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

neela | High School Teacher | (Level 3) Valedictorian

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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 .

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll put the equtaion 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