The endpoint of its diameter is (1,0) and (-3,4) , find the equation of the circle?

Question

how Find the equation of circle which passes through the points A(1,3) B(3,-1) C(4,0)? too?

Answer 1

(1).

The endpoints of the diameter are (1, 0) and (-3, 4).

The center of the circle = The midpoint of the two points (1, 0) and (-3, 4)

                                     =[(1-3)/2, (0+4)/2]

                                     = (-1, 2).

Radius of the circle = The distance between two points (1, 0) and (-1, 2).

                               = sqrt[(2-0)^2+(-1-1)^2]

                               = sqrt[4+4]

                               = sqrt[8].

Standard form of circle equation : (x - h)^2 + (y - k)^2 = r^2, where (h, k) = center and t = radius.

[x - (-1)]^2 + [y - (2)]^2 = 8

(x + 1)^2 + (y - 2)^2 = 8.

Answer 2

(2).

The points on the circle are A(1, 3), B(3, -1) and C(4, 0).

The general equation of the circle is x^2 + y^2 +2gx +2fy + c = 0.

A(1, 3) ---> (1)^2 + (3)^2 +2g(1) +2f(3) + c = 0 ------> 2g +6f + c = -10 -----> (1).

B(3, -1) ---> (3)^2 + (-1)^2 +2g(3) +2f(-1) + c = 0 ------> 6g - 2f + c = -10 ------> (2).

(4, 0) ---> (4)^2 + (0)^2 +2g(4) +2f(0) + c = 0 ----> 8g + c = -16 ----> c = -16 - 8g ----> (3).

Equation 1: 2g +6f + (-16 - 8g) = -10 -----> - 6g +6f = 6 -----> - g + f = 1 -----> (4).

Equation 2: 6g - 2f + (-16 - 8g) = -10 -----> - 2g - 2f = 6 ----->  g + f = -3 -----> (5).

Add equation 4 and equation 5 -----> 2f = -2 ---> f = -2.

From equation 4: - g + (-2) = 1 -----> g = -3

From equation 3: c = -16 - 8(-3) -----> c = 8.

The circle equation is x^2 + y^2 -6x -4y + 8 = 0.