I need to write the standard equation for the circle that passes through the points:
(-5,0) (0,4) (2,4)
I have to use the general equation for a circle:
x^2+y^2+dx+ey+f=0
The center will lie at the intersection of the perpendicular bisectors of the line segments (chords) joining the points.
If we label the points A,B,C, then
The midpoint of AB is at (-5/2,2)
The slope of the chord AB is 4/5
The line AB is y = 4/5 (x+5)
The slope of the radius to AB is thus -5/4
The equation of the radius to AB is y = -5/4(x+5/2)+2
The midpoint of BC is at (1,4)
The slope of the chord BC is 0
The line BC is y=4
The radius to BC is vertical
The equation of the radius to BC is x=1
The midpoint of AC is at (-3/2,2)
The slope of the chord AC is 4/7
The line AC is y = 4/7(x+5)
The slope of the radius to AC is thus -7/4
The equation of the radius to AC is y = -7/4(x+3/2)+2
The radii intersect at (1,-19/8), which is the center of the circle.
The distance from the center to (0,4) is r^2 = 1+2601/64 = 2665/64
So, the circle is
(x-1)^2 + (y + 19/8)^2 = 2665/64
You can see the circle and the chords at
http://www.wolframalpha.com/input/?i=plot+%28x-1%29^2+%2B+%28y+%2B+19%2F8%29^2+%3D+2665%2F64%2C+y+%3D+4%2F5+%28x%2B5%29%2C++y%3D4%2C+y+%3D+4%2F7%28x%2B5%29&dataset=&equal=Submit
sketch a graph first
If it goes through (0,4) and (2,4) I suspect that the center is on the line x = 1
(x-1)^2 + (y-b)^2 = r^2
when x = 0, y = 4
1 + (4-b)^2 = r^2
when x = 2, y = 4
1 + (4-b)^2 = r^2 again of course
when x = -5, y = 0
36 + b^2 = r^2
so
1 +(4-b)^2 = b^2+36
1 + 16 -8b + b^2 = b^2 + 36
8 b = 17 - 36 = -19
b = -19/8
so
(x-1)^2 + (y+19/8)^2 = 36 + 361/64 = 2665/64
(x-1)^2 + (y+19/8)^2 = 6.45^2
check my arithmetic and you can put that in your form if you choose
Or, knowing that the general circle is
(x-h)^2 + (y-k)^2 = r^2
plug in your points to get
(-5-h)^2 + k^2 = r^2
h^2 + (4-k)^2 = r^2
(2-h)^2 + (4-k)^2 = r^2
h = 1
k = -19/8
r^2 = 2665/64
and we have the circle as derived above
The brute-force way, of course, is to plug your points into
x^2+y^2+dx+ey+f=0
(-5,0): 25-5d+f = 0
(0,4): 16+4e+f = 0
(2,4): 4+16+2d+4e+f = 0
d,e,f = -2, 19/4, -35
and the circle is thus
x^2 + y^2 - 2x + 19/4 y - 35 = 0
I'll let you verify that this is the same circle.
Let the circle pass through, in general, three points:
(x1,y1)
(x2,y2)
(x3,y3)
and the equation is to be expressed in the form
x²+y²+dx+ey+f=0
The required equation may be expressed in determinant form:
[y^2+x^2, x, y, 1]
[y1^2+x1^2, x1, y1, 1]
[y2^2+x2^2, x2, y2, 1]
[y3^2+x3^2, x3, y3, 1]
When the determinant is expanded and equated to zero, the equation of the given circle in the required (polynomial) form will result.
For the given three points (-5,0),(0,4),(2,4),
the determinant expands to:
x²+y²-2x+(19/4)y-35=0
This is more fun every hour! Four ways now :)
To find the standard equation for a circle, you need to follow the steps:
Step 1: Find the center of the circle.
Step 2: Use the center and any given point to find the radius.
Step 3: Substitute the values into the general equation for a circle.
Let's go through each step:
Step 1: Find the center of the circle.
To find the center, you can use the formula: (h, k), where h is the x-coordinate of the center and k is the y-coordinate of the center.
To find the center, you can take the average of the x-coordinates and the average of the y-coordinates of the given points:
Average of x-coordinates: (-5 + 0 + 2)/3 = -3/3 = -1
Average of y-coordinates: (0 + 4 + 4)/3 = 8/3 ≈ 2.67
Therefore, the center of the circle is (-1, 2.67).
Step 2: Use the center and any given point to find the radius.
The radius can be found using the distance formula: √((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) is the center and (x2, y2) is any given point on the circle.
Let's take the first given point (-5, 0):
Distance = √((-5 - (-1))^2 + (0 - 2.67)^2) = √((-5 + 1)^2 + (-2.67)^2) = √((-4)^2 + (-2.67)^2) = √(16 + 7.1289) = √23.1289 ≈ 4.809
Therefore, the radius of the circle is approximately 4.809.
Step 3: Substitute the values into the general equation for a circle.
The general equation for a circle is: x^2 + y^2 + dx + ey + f = 0
Using the values we found, substitute them into the equation:
(x - h)^2 + (y - k)^2 = r^2
(x - (-1))^2 + (y - 2.67)^2 = (4.809)^2
(x + 1)^2 + (y - 2.67)^2 = 23.1289
Expanding this equation will give you the standard equation for the circle that passes through the given points:
x^2 + 2x + 1 + y^2 - 5.34y + 7.1289 = 23.1289
x^2 + y^2 + 2x - 5.34y - 15 = 0
Therefore, the standard equation for the circle that passes through the points (-5, 0), (0, 4), and (2, 4) is x^2 + y^2 + 2x - 5.34y - 15 = 0.