Please help find the equation of the circle where (−4, −4) and (12, 14) is a diameter.

Find the equation of the circle for which the line segment determined by (−4, −4) and (12, 14) is a diameter.(Please express the final equation in the form x2 + y2 + Dx + Ey + F = 0.)

1st we solve the mid point of the line of A and B in the center of a circle..
........x1 + x2..........y1 + y2
x = ---------------, y = -------------
..........2......................2

.......- 4 + 12............- 4 + 14
h = ---------------, k = ----------------
............2......................2

h = 4, k = 5

the center of the circle is C(4,5)

now solving for the radius using the distance formula..


r = √[(x - h)^2 + (y - k)^2]
r = √[(- 4 - 4)^2 + (- 4 - 5)^2]
r = √[64 + 81]
r = √(145)

an equation of the circle is therefore

(x - h)^2 + (y - k)^2 = r^2
(x - 4)^2 + (y - 5)^2 = √(145)^2
x^2 - 8x + 16 + y^2 - 10y + 25 = 145
x^2 + y^2 - 8x - 10y + 16 + 25 - 145 = 0
x^2 + y^2 - 8x - 10y - 104 = 0 answer//

Midpoint diameter = (4,5)
Therefore, equation is (x-4)^2 + (y-5)^2 = r^2
Radius is (4,5) to (12,14). Use pythagoras to find length. 8^2 + 9^2 = 64 + 81 = 145... so radius is sqrt145
New equation is (x-4)^2 + (y-5)^2 = 145
Expand:
x^2 - 8x + 16 + y^2 - 10y +25 = 145
Collect altogether:
x^2 + y^2 - 8x - 10y - 104 = 0

If (-4, -4) and (12, 14) are the endpoints of the diameter, then (4, 5) is the center.

√((12 - 4)^2 + (14 - 5)^2) = √(10^2 + 9^2) = √181 is the radius

(x - 4)^2 + (y - 5)^2 = 181
x^2 - 8x + 16 + y^2 - 10y + 25 = 181
x^2 = y^2 - 8x - 10y - 140 = 0 **