Question 36377: What is the y-coordinate of the center of the circle that passes through (-1,2),
(3,2), and (5,4)?
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website! What is the y-coordinate of the center of the circle that
passes through (-1,2), (3,2), and (5,4)?
The general equation of a circle is
x² + y² + Dx + Ey + F = 0
Substituting first point, (-1,2), i.e., x = -1, y = 2
(-1)² + (2)² + D(-1) + E(2) + F = 0
1 + 4 - D + 2E + F = 0
5 - D + 2E + F = 0
-D + 2E + F = -5
Substituting second point, (3,2), i.e., x = 3, y = 2
(3)² + (2)² + D(3) + E(2) + F = 0
9 + 4 + 3D + 2E + F = 0
13 + 3D + 2E + F = 0
3D + 2E + F = -13
Substituting third point, (5,4), i.e., x = 5, y = 4
(5)² + (4)² + D(5) + E(4) + F = 0
25 + 16 + 5D + 4E + F = 0
41 + 5D + 4E + F = 0
5D + 4E + F = -41
That gives you a system of three equations in
three unknowns:
-D + 2E + F = -5
3D + 2E + F = -13
5D + 4E + F = -41
D = -2, E = -12, F = 17
The general form the equation of the circle is
x² + y² - 2x - 12y + 17 = 0
Now we must get it in standard form
(x - h)² + (y - k)² = r²
x² + y² - 2x - 12y + 17 = 0
x² - 2x + y² - 12y = -17
(x² - 2x) + (y² - 12y) = -17
Complete squares:
Multiply coeff's of x and y by 1/2, square the results,
add inside parentheses and also to right side.
(x² - 2x + 1) + (y² - 12y + 36) = -17 + 1 + 36
Factor the trinomials in parentheses, combine numbers
on right:
(x - 1)² + (y - 6)² = 20
Compare to (x - h)² + (y - k)² = r²
h = 1, k = 6, r² = 20, so center is (1, 6),
__ _
radius = r = Ö20 = 2Ö5
You were asked only for the y-coordinate of the center,
which is k = 6
Edwin McCravy
AnlytcPhil@aol.com
|
|
|
