Question

Find the equation of a circle with the Endpoints of a diameter: (11, −5), (3, 15)

Answer 1

[x-7]^2 + [y-5]^2= 116

Explanation:

General equation of a circle is

[x2 - xo]^2 + [y2- yo]^2 = r^2

Where xo and yo are coordinates of the circle at the origin.

But xo = [x2 + x1]/2

= 11 + 3 / 2 = 7

yo = [y2 + y1] / 2 = [-5 + 15] /2 = 5

x2= 11, x2= 3; y1= -5, y2= 15

[11-7]^2 + [15-5]^2 =

16+ 100=116 = r^2

From the expression below;

[x2-xo]^2 + [y2-yo] ^2 = r^2

[x1-xo]^2 + [y1-yo] ^2 = r^2

[x-7]^2 + [y-5]^2= 116

Answer 2

(x - 7)² + (y - 5)² = 116

Explanation:

equation of a circle:

(x - h)² + (y - k)² = r²

where (h,k) is the centre and r is the radius

centre is the midpoint of the diameter:

(h,k) = (11+3)/2 , (-5+15)/2

(h,k) = (7,5)

diameter = sqrt[(15--5)² + (3-11)²]

diameter = sqrt(464) = 4sqrt(29)

radius = ½ diameter

radius = 2sqrt(29)

r² = 116

equation:

(x - 7)² + (y - 5)² = 116