the circle passes through the point (0,0), (5,0) and (3,3)
You can do this either by using the general form or the standard form.
I'll first do it with the general form:
x² + y² + Dx + Ey + F = 0
Substitute (x,y) = (0,0)
0² + 0² + D(0) + E(0) + F = 0
F = 0
So we have
x² + y² + Dx + Ey + 0 = 0
x² + y² + Dx + Ey = 0
Substitute (x,y) = (5,0)
5² + 0² + D(5) + E(0) = 0
25 + 5D = 0
25 + 5D = 0
5D = -25
D = -5
So we have:
x² + y² + Dx + Ey = 0
x² + y² - 5x + Ey = 0
Substitute (x,y) = (3,3)
x² + y² - 5x + Ey = 0
(3)² + (3)² - 5(3) + E(3) = 0
9 + 9 - 15 + 3E = 0
3 + 3E = 0
3E = -3
E = -1
x² + y² - 5x + Ey = 0
x² + y² - 5x - y = 0
------------------------------------------
You can also use the standard form equation for a circle:
(x - h)² + (y - k)² = r²
Substitute (x,y) = (0,0)
(0 - h)² + (0 - k)² = r²
h² + k² = r²
Substitute (x,y) = (5,0)
(5 - h)² + (0 - k)² = r²
25 - 10h + h² + k² = r²
Substitute (x,y) = (3,3)
(3 - h)² + (3 - k)² = r²
9 - 6h + h² + 9 - 6k + k² = r²
18 - 6h - 6k + h² + k² = r²
So we have this system of equations:
h² + k² = r²
25 - 10h + h² + k² = r²
18 - 6h - 6k + h² + k² = r²
If we subtract the first equation term by term from each of the
other two equations we have:
25 - 10h = 0
18 - 6h - 6k = 0
Solving the first for h
25 - 10h = 0
-10h = -25
h =
h =
Substituting in
18 - 6h - 6k = 0
18 - 6 - 6k = 0
18 - 15 - 6k = 0
3 - 6k = 0
-6k = -3
k =
k =
Substituting in
h² + k² = r²
So the equation
(x - h)² + (y - k)² = r²
becomes
(x - )² + (y - )² =
It was harder but mainly because the center and radius were fractions.
Here is the graph. Notice that the circle passes through the
given points and has center (,) or (2.5,.5)
The radius is ≈ 2.55
Edwin