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Find the equation of a circle that passes through the points (2, 3), (6, 1), and (4, -3).
Solution:
To illustrate the problem, it is better to sketch the circle with three points as follows:
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| Photo by Math Principles in Everyday Life |
We know that the equation of a circle in standard form is
(x - h)2 + (y - k)2 = r2
where (h, k) is the coordinate of the center of a circle and r is the radius of a circle. In this problem, the center and radius of a circle are not given and we cannot use the equation of a circle in standard form.
The equation of a circle in general form is
x2 + y2 + Dx + Ey + F = 0
where D, E, and F are the coefficients. Since the three points of a circle are given, we can use the equation of a circle in general form to solve for D, E, and F.
For (2, 3), if x = 2 and y = 3, the equation becomes
(2)2 + (3)2 + D(2) + E(3) + F = 0
4 + 9 + 2D + 3E + F = 0
2D + 3E + F = -13 (equation 1)
For (6, 1), if x = 6 and y = 1, the equation becomes
(6)2 + (1)2 + D(6) + E(1) + F = 0
36 + 1 + 6D + E + F = 0
6D + E + F = -37 (equation 2)
For (4, -3), if x = 4 and y = -3, the equation becomes
(4)2 + (-3)2 + D(4) + E(-3) + F = 0
16 + 9 + 4D - 3E + F = 0
4D - 3E + F = -25 (equation 3)
Since the three points of a circle are given, there are three equations and three unknowns. We can now solve for D, E, and F by elimination method.
If you subtract equation 2 from equation 1,
2D + 3E + F = -13 2D + 3E + F = -13
--------->
-(6D + E + F = -37) -6D - E - F = 37
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-4D + 2E = 24
-2D + E = 12 (equation 4)
If you subtract equation 3 from equation 2,
6D + E + F = -37 6D + E + F = -37
--------->
-(4D - 3E + F = -25) -4D + 3E - F = 25
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2D + 4E = - 12 (equation 5)
Add equation 4 and equation 5,
-2D + E = 12
2D + 4E = -12
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5E = 0
E = 0
Substitute the value of E to equation 4 or equation 5,
2D + 4E = -12
2D + 4(0) = -12
2D = -12
D = -6
Substitute the value of D and E to equation 1, equation 2, or equation 3,
2D + 3E + F = -13
2(-6) + 3(0) + F = -13
-12 + F = -13
F = -1
Substitute the value of D, E, and F to the equation of a circle in general form,
x2 + y2 + Dx + Ey + F = 0
x2 + y2 + (-6)x + (0)y + (-1) = 0
Therefore, the equation of a circle is
x2 + y2 - 6x - 1 = 0