FINDING CENTER AND RADIUS OF A CIRCLE BY COMPLETING THE SQUARE
Sometimes equation of circle will be in the form
x2 + y2 + 2gx + 2fy + c = 0
Using completing the square method, we can get the equation in the form
(x - h)2 + (y - k)2 = r2
Here (h, k) is the center and r is the radius.
Using the algebraic identities,
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
Use the information provided to write the standard form equation of each circle.
Problem 1 :
x2 + y2 + 14x - 12y + 4 = 0
Solution:
x2 + y2 + 14x - 12y + 4 = 0
x2 + 14x + y2 - 12y + 4 = 0
x2 + 2(x)(7) + y2 - 2(y)(6) + 4 = 0
x2 + 2(x)(7) + 72 - 72 + y2 - 2(y)(6) + 62 - 62 + 4 = 0
Using the expansions of (a - b)2 and (a + b)2,
(x + 7)2 - 72 + (y - 6)2 - 62 + 4 = 0
(x + 7)2 - 49 + (y - 6)2 - 36 + 4 = 0
(x + 7)2 + (y - 6)2 - 81 = 0
(x + 7)2 + (y - 6)2 = 81
(x + 7)2 + (y - 6)2 = 92
Comparing with (x - h)2 + (y - k)2 = r2
Center (h, k) is (-7, 6) and radius is 9.
Problem 2 :
y2 + 2x + x2 = 24y - 120
Solution:
y2 + 2x + x2 = 24y - 120
x2 + 2x + y2 - 24y + 120 = 0
x2 + 2(x)(1) + y2 - 2(y)(12) + 120 = 0
x2 + 2(x)(1) + 12 - 12 + y2 - 2(y)(12) + 122 - 122 + 120 = 0
Using the expansions of (a - b)2 and (a + b)2,
(x + 1)2 - 12 + (y - 12)2 - 122 + 120 = 0
(x + 1)2 - 1 + (y - 12)2 - 144 + 120 = 0
(x + 1)2 + (y - 12)2 - 25 = 0
(x + 1)2 + (y - 12)2 = 25
Comparing with (x - h)2 + (y - k)2 = r2
Center (h, k) is (-1, 12) and radius is 5.
Problem 3 :
8x + x2 - 2y = 64 - y2
Solution:
8x + x2 - 2y = 64 - y2
x2 + 8x + y2 - 2y - 64 = 0
x2 + 2(x)(4) + y2 - 2(y)(1) - 64 = 0
x2 + 2(x)(4) + 42 - 42 + y2 - 2(y)(1) + 12 - 12 - 64 = 0
Using the expansions of (a - b)2 and (a + b)2,
(x + 4)2 - 42 + (y - 1)2 - 12 - 64 = 0
(x + 4)2 - 16 + (y - 1)2 - 1 - 64 = 0
(x + 4)2 + (y - 1)2 - 81 = 0
(x + 4)2 + (y - 1)2 = 81
(x + 4)2 + (y - 1)2 = 92
Comparing with (x - h)2 + (y - k)2 = r2
Center (h, k) is (-4, 1) and radius is 9
Problem 4 :
137 + 6y = -y2 - x2 - 24x
Solution:
137 + 6y = -y2 - x2 - 24x
x2 + 24x + y2 + 6y + 137 = 0
x2 + 2(x)(12) + y2 + 2(y)(3) + 137 = 0
x2 + 2(x)(12) + 122 - 122 + y2 + 2(y)(3) + 32 - 32 + 137 = 0
Using the expansions of (a - b)2 and (a + b)2,
(x + 12)2 - 122 + (y + 3)2 - 32 + 137 = 0
(x + 12)2 - 144 + (y + 3)2 - 9 + 137 = 0
(x + 12)2 + (y + 3)2 - 16 = 0
(x + 12)2 + (y + 3)2 = 16
(x + 12)2 + (y + 3)2 = 42
Comparing with (x - h)2 + (y - k)2 = r2
Center (h, k) is (-12, -3) and radius is 4.
Problem 5 :
x2 + 2x + y2 = 55 + 10y
Solution:
x2 + 2x + y2 = 55 + 10y
x2 + 2x + y2 - 10y - 55 = 0
x2 + 2(x)(1) + y2 - 2(y)(5) - 55 = 0
x2 + 2(x)(1) + 12 - 12 + y2 - 2(y)(5) + 52 - 52 - 55 = 0
Using the expansions of (a - b)2 and (a + b)2,
(x + 1)2 - 12 + (y - 5)2 - 52 - 55 = 0
(x + 1)2 - 1 + (y - 5)2 - 25 - 55 = 0
(x + 1)2 + (y - 5)2 - 81 = 0
(x + 1)2 + (y - 5)2 = 81
Comparing with (x - h)2 + (y - k)2 = r2
Center (h, k) is (-1, 5) and radius is 9.
Problem 6 :
8x + 32y + y2 = -263 - x2
Solution:
8x + 32y + y2 = -263 - x2
x2 + 8x + y2 + 32y + 263 = 0
x2 + 2(x)(4) + y2 + 2(y)(16) + 263 = 0
x2 + 2(x)(4) + 42 - 42 + y2 + 2(y)(16) + 162 - 162 + 263 = 0
Using the expansions of (a - b)2 and (a + b)2,
(x + 4)2 - 42 + (y + 16)2 - 162 + 263 = 0
(x + 4)2 - 16 + (y + 16)2 - 256 + 263 = 0
(x + 4)2 + (y + 16)2 - 9 = 0
(x + 4)2 + (y + 16)2 = 9
Comparing with (x - h)2 + (y - k)2 = r2
Center (h, k) is (-4, -16) and radius is 3.
Recent Articles
-
Finding Range of Values Inequality Problems
May 21, 24 08:51 PM
Finding Range of Values Inequality Problems -
Solving Two Step Inequality Word Problems
May 21, 24 08:51 AM
Solving Two Step Inequality Word Problems -
Exponential Function Context and Data Modeling
May 20, 24 10:45 PM
Exponential Function Context and Data Modeling