Find an equation of the circle that satisfies the given conditions.

Algebra
Tutor: None Selected Time limit: 1 Day

Endpoints of a diameter are 

P(−2, 1) and Q(4, −5).
Jan 29th, 2015

Endpoints of a diameter:  P(-2, 1) and Q(4, -5)

Diameter: d = sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2]

d = sqrt[(4 - -2)^2 + (-5 - 1)^2]

d = sqrt[(6)^2 + (-6)^2] = sqrt[36 + 36]

d = sqrt[72] = sqrt[9  * 8] = 3 * 2sqrt(2) = 6 * sqrt(2)

Radius:

r = d/2 = [6 * sqrt(2)]/2 = 3 * sqrt(2)

Center of Circle:

h = (x_1 + x_2)/2 = (-2 + 4)/2 = 1

k = (y_1 + y_2)/2 = (1 + -5)/2 = -4/2 = -2

Center = (h, k) = (1, -2)

Equation of Circle:

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

(x - 1)^2 + (y - (-2))^2 = (3 * sqrt(2))^2

(x - 1)^2 + (y + 2)^2 = 18


Answer:  The equation of the circle that fits the given conditions is (x - 1)^2 +  (y + 2)^2 = 18

Jan 29th, 2015

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