Problem 1 :
Find the coordinates of the center of a circle passing through the points A(0, 0), B(-2, 1) and C(-3, 2). Also find the radius of this circle.
Solution:
The general equation of a circle
x^{2} + y^{2} + 2gx + 2fy + c = 0
Equation of a circle which is passing through the point (0, 0).
0 + 0 + 2g(0) + 2f(0) + c = 0
c = 0
Equation of a circle which is passing through the point (-2, 1).
(-2)^{2} + (1)^{2} + 2g(-2) + 2f(1) + c = 0
5 - 4g + 2f + c = 0
-4g + 2f + c = -5 ----> (1)
Equation of a circle which is passing through the point (-3, 2).
(-3)^{2} + (2)^{2} + 2g(-3) + 2f(2) + c = 0
13 - 6g + 4f + c = 0
-6g + 4f + c = -13 ----> (2)
By applying c = 0 in (1) and (2)
-4g + 2f = -5 ---> (3)
-6g + 4f = -13 ---> (4)
(1) × 2 ==> -8g + 4f = -10 ----> (5)
Subtract (4) from this equation
-8g + 6g + 4f - 4f = -10 + 13
-2g = 3
g = -3/2
Substitute g = -3/2 in equation (3)
-4(-3/2) + 2f = -5
6 + 2f = -5
2f = -11
f = -11/2
Substitute g = -3/2, f = -11/2 and c = 0 in the general equation
x^{2} + y^{2} + 2gx + 2fy + c = 0
x^{2} + y^{2} + 2(-3/2)x + 2(-11/2)y + 0 = 0
x^{2} + y^{2} - 3x - 11y = 0
So, the equation of a circle passing through three points
x^{2} + y^{2} - 3x - 11y = 0
Problem 2 :
Find the center of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Solution:
The general equation of a circle
x^{2} + y^{2} + 2gx + 2fy + c = 0
Equation of a circle which is passing through the point (6, -6).
(6)^{2} + (-6)^{2} + 2g(6) + 2f(-6) + c = 0
36 + 36 + 12g - 12f + c = 0
72 + 12g - 12f + c = 0
12g - 12f + c = -72 ---> (1)
Equation of a circle which is passing through the point (3, -7).
(3)^{2} + (-7)^{2} + 2g(3) + 2f(-7) + c = 0
9 + 49 + 6g - 14f + c = 0
58 + 6g - 14f + c = 0
6g - 14f + c = -58 ---> (2)
Equation of a circle which is passing through the point (3, 3).
(3)^{2} + (3)^{2} + 2g(3) + 2f(3) + c = 0
9 + 9 + 6g + 6f + c = 0
18 + 6g + 6f + c = 0
6g + 6f + c = -18 ---> (3)
So, the three equations are
12g - 12f + c = -72 ---> (1)
6g - 14f + c = -58 ---> (2)
6g + 6f + c = -18 ---> (3)
By solving these three equations we can get the values of f, g and c.
By adding (1) and (2),
6g + 2f = -14 ---> (4)
By subtracting (2) and (3)
(6g - 14f + c) - (6g + 6f + c) = -58 + 18
-20f = -40
f = 2
Substitute f = 2 in equation (4),
6g + 2(2) = -14
6g + 4 = -14
6g = -18
g = -3
Substitute f = 2 and g = -3 in equation (1)
12g - 12f + c = -72
12(-3) - 12(2) + c = -72
-36 - 24 + c = -72
-60 + c = -72
c = -72 + 60
c = -12
Substitute g = -3, f = 2 and c = -12 in the general equation
x^{2} + y^{2} + 2gx + 2fy + c = 0
x^{2} + y^{2} + 2(-3)x + 2(2)y - 12 = 0
x^{2} + y^{2} - 6x + 4y - 12= 0
So, the equation of a circle passing through three points
x^{2} + y^{2} - 6x + 4y - 12 = 0
Problem 3 :
Find the coordinates of the point equidistant from three given points A(5, 3), B(5, -5) and C(1, -5).
Solution:
The general equation of a circle
x^{2} + y^{2} + 2gx + 2fy + c = 0
Equation of a circle which is passing through the point (5, 3).
(5)^{2} + (3)^{2} + 2g(5) + 2f(3) + c = 0
25 + 9 + 10g + 6f + c = 0
34 + 10g + 6f + c = 0
10g + 6f + c = -34 ---> (1)
Equation of a circle which is passing through the point (5, -5).
(5)^{2} + (-5)^{2} + 2g(5) + 2f(-5) + c = 0
25 + 25 + 10g - 10f + c = 0
50 + 10g - 10f + c = 0
10g - 10f + c = -50 ---> (2)
Equation of a circle which is passing through the point (1, -5).
(1)^{2} + (-5)^{2} + 2g(1) + 2f(-5) + c = 0
1 + 25 + 2g - 10f + c = 0
26 + 2g - 10f + c = 0
2g - 10f + c = -26 ---> (3)
So, the three equations are
10g + 6f + c = -34 ---> (1)
10g - 10f + c = -50 ---> (2)
2g - 10f + c = -26 ---> (3)
By subtracting (1) and (2)
(10g + 6f + c) - (10g - 10f + c) = -34 + 50
16f = 16
f = 1
By subtracting (2) and (3)
(10g - 10f + c) - (2g - 10f + c) = -50 + 26
8g = -24
g = -3
Substitute f = 1 and g = -3 in equation (1)
10(-3) + 6(1) + c = -34
-30 + 6 + c = -34
-24 + c = -34
c = -10
Substitute g = -3, f = 1 and c = -10 in the general equation
x^{2} + y^{2} + 2gx + 2fy + c = 0
x^{2} + y^{2} + 2(-3)x + 2(1)y - 10 = 0
x^{2} + y^{2} - 6x + 2y - 10 = 0
So, the equation of a circle passing through three points
x^{2} + y^{2} - 6x + 2y - 10 = 0.
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