The base equation for a circle centered at (h,k) with radius r is:
(x - h)2+ (y - k)2 = r2
If the points are concyclic, they all lie on the graph for the same circle.
Thus, all points are solutions to a single equation of the circle
WE can plug each point into the base equation, and obtain a system of equationd
FOR POINT (-2,10)
Equation 1: (-2 - h)2 + (10 - k)2 = r2
FOR POINT (1,11)
Equation 2: (1 - h)2+ (11 - k)2 = r2
FOR POINT (6,10)
Equation 3: (6 - h)2 + (10 - k)2 = r2
FOR POINT (9,7)
Equation 4: (9 - h)2 + (7 - k)2 = r2
From Equation 1 and 2 above, since they have 2 different expressions equivalent to r2, we can construct an equation involving variables x and h.
r2 = r2
(-2 - h)2 + (10 - k)2 = (1 - h)2 + (11 - k)2
(-2 - h)(-2 - h) + (10 - k)(10 - k) = (1 - h)(1 - h) + (11 - k)(11 - k)
(4 + 2h + 2h + h2) + (100 -10k - 10k + k2) = (1 - h - h + h2) + (121 - 11k - 11k + k2)
h2 + 4h + 4 + k2 - 20k + 100 = h2-2h + 1 + k2 -22k + 121
We can subtract h and k from both sides.
4h + 4 - 20k + 100 = -2h + 1 -22k + 122
4h -20k + 104 = -2h - 22k + 122
Move all h and k variables to the left ... move all constants to the right
6H + 2K = 18
SYSTEM EQUATION 1: 6h + 2k = 18
From equation 3 and 4 above, we go through the same process to obtain a second system equation.
r2 = r2
(6 - h)2 + (10 - k)2 = (9 - h)2 + (7 - k)2
(6 - h)(6 - h) + (10 - k)(10 - k) = (9 - h)(9 - h) + (7 - k)(7 - k)
36 - 6h - 6h + h2+ 100 - 10k - 10k + k2 = 81 - 9h - 9h + h2 + 49 = 7k - 7k + k2
h2 - 12h + 36 + k2 - 20k + 100 = h2 - 18h + 81 + k2- 14k + 49
Simplify, and subtract h2 and k2from both sides
-12h - 20k + 136 = -18h - 14k + 130
Move the h and k variables to the left, and the constants to the right
-12h + 18h - 20k + 14k = 130 - 136
6h - 6k = -6
SYSTEM EQUATION 1: 6h - 6k = -6
6h +2k = 18
6h - 6k = -6
Multiply the second equation by -1, we have ... -6h + 6k = 6
Adding the 2 equations, we have
6h + 2k = 18
-6h + 6k = 6
8k =24
k = 3
Using substitution of k=3 into equation 6h + 2k = 18,
6h + 2(3) = 18
6h + 6 = 18
6h = 12
h = 2
Now, use one of the points on the circle ... (1, 11) and (h,k) ... (2,3), to find the radius
(1 - 2)2 + (11 - 3)2 = r2
1 + 64 = r2
65 = r2
r = √65
Thus the equation of our circle is:
(x - 2)2 + (y - 3)2 = (√65)2
Yogita Ingle 4 years, 3 months ago
The base equation for a circle centered at (h,k) with radius r is:
(x - h)2+ (y - k)2 = r2
If the points are concyclic, they all lie on the graph for the same circle.
Thus, all points are solutions to a single equation of the circle
WE can plug each point into the base equation, and obtain a system of equationd
FOR POINT (-2,10)
Equation 1: (-2 - h)2 + (10 - k)2 = r2
FOR POINT (1,11)
Equation 2: (1 - h)2+ (11 - k)2 = r2
FOR POINT (6,10)
Equation 3: (6 - h)2 + (10 - k)2 = r2
FOR POINT (9,7)
Equation 4: (9 - h)2 + (7 - k)2 = r2
From Equation 1 and 2 above, since they have 2 different expressions equivalent to r2, we can construct an equation involving variables x and h.
r2 = r2
(-2 - h)2 + (10 - k)2 = (1 - h)2 + (11 - k)2
(-2 - h)(-2 - h) + (10 - k)(10 - k) = (1 - h)(1 - h) + (11 - k)(11 - k)
(4 + 2h + 2h + h2) + (100 -10k - 10k + k2) = (1 - h - h + h2) + (121 - 11k - 11k + k2)
h2 + 4h + 4 + k2 - 20k + 100 = h2-2h + 1 + k2 -22k + 121
We can subtract h and k from both sides.
4h + 4 - 20k + 100 = -2h + 1 -22k + 122
4h -20k + 104 = -2h - 22k + 122
Move all h and k variables to the left ... move all constants to the right
6H + 2K = 18
SYSTEM EQUATION 1: 6h + 2k = 18
From equation 3 and 4 above, we go through the same process to obtain a second system equation.
r2 = r2
(6 - h)2 + (10 - k)2 = (9 - h)2 + (7 - k)2
(6 - h)(6 - h) + (10 - k)(10 - k) = (9 - h)(9 - h) + (7 - k)(7 - k)
36 - 6h - 6h + h2+ 100 - 10k - 10k + k2 = 81 - 9h - 9h + h2 + 49 = 7k - 7k + k2
h2 - 12h + 36 + k2 - 20k + 100 = h2 - 18h + 81 + k2- 14k + 49
Simplify, and subtract h2 and k2from both sides
-12h - 20k + 136 = -18h - 14k + 130
Move the h and k variables to the left, and the constants to the right
-12h + 18h - 20k + 14k = 130 - 136
6h - 6k = -6
SYSTEM EQUATION 1: 6h - 6k = -6
6h +2k = 18
6h - 6k = -6
Multiply the second equation by -1, we have ... -6h + 6k = 6
Adding the 2 equations, we have
6h + 2k = 18
-6h + 6k = 6
8k =24
k = 3
Using substitution of k=3 into equation 6h + 2k = 18,
6h + 2(3) = 18
6h + 6 = 18
6h = 12
h = 2
Now, use one of the points on the circle ... (1, 11) and (h,k) ... (2,3), to find the radius
(1 - 2)2 + (11 - 3)2 = r2
1 + 64 = r2
65 = r2
r = √65
Thus the equation of our circle is:
(x - 2)2 + (y - 3)2 = (√65)2
0Thank You