Finding the Equation of a Circle Passing Through Three Points

To find the equation of a circle that passes through three given points, we can use the general form of the equation of a circle:

\( x^2 + y^2 + 2gx + 2fy + c = 0 \)

where \((-g, -f)\) is the center of the circle and \( \sqrt{g^2 + f^2 - c} \) is the radius. We have three unknown constants: \(g\), \(f\), and \(c\). Since the circle passes through the three given points, each point must satisfy the equation. We can substitute the coordinates of each point into the general equation to form a system of three linear equations in terms of \(g\), \(f\), and \(c\).

Substituting the Given Points

The given points are (1, 0), (0, -6), and (3, 4).

1. Point (1, 0): Substituting \(x=1\) and \(y=0\) into the general equation: \( (1)^2 + (0)^2 + 2g(1) + 2f(0) + c = 0 \) \( 1 + 0 + 2g + 0 + c = 0 \) \( 2g + c = -1 \) (Equation 1)
2. Point (0, -6): Substituting \(x=0\) and \(y=-6\) into the general equation: \( (0)^2 + (-6)^2 + 2g(0) + 2f(-6) + c = 0 \) \( 0 + 36 + 0 - 12f + c = 0 \) \( -12f + c = -36 \) (Equation 2)
3. Point (3, 4): Substituting \(x=3\) and \(y=4\) into the general equation: \( (3)^2 + (4)^2 + 2g(3) + 2f(4) + c = 0 \) \( 9 + 16 + 6g + 8f + c = 0 \) \( 25 + 6g + 8f + c = 0 \) \( 6g + 8f + c = -25 \) (Equation 3)

Solving the System of Equations

Now we have a system of three linear equations:

1. \( 2g + c = -1 \)
2. \( -12f + c = -36 \)
3. \( 6g + 8f + c = -25 \)

We can solve this system to find the values of \(g\), \(f\), and \(c\).

From Equation 1, we can express \(c\) in terms of \(g\):

\( c = -1 - 2g \)

Substitute this expression for \(c\) into Equation 2:

\( -12f + (-1 - 2g) = -36 \)

\( -12f - 1 - 2g = -36 \)

\( -2g - 12f = -35 \)

\( 2g + 12f = 35 \) (Equation 4)

Substitute the expression for \(c\) into Equation 3:

\( 6g + 8f + (-1 - 2g) = -25 \)

\( 6g + 8f - 1 - 2g = -25 \)

\( 4g + 8f = -24 \)

Divide by 4:

\( g + 2f = -6 \) (Equation 5)

Now we have a system of two equations with two variables, \(g\) and \(f\):

1. \( 2g + 12f = 35 \)
2. \( g + 2f = -6 \)

From Equation 5, we can express \(g\) in terms of \(f\):

\( g = -6 - 2f \)

Substitute this expression for \(g\) into Equation 4:

\( 2(-6 - 2f) + 12f = 35 \)

\( -12 - 4f + 12f = 35 \)

\( 8f = 35 + 12 \)

\( 8f = 47 \)

\( f = \frac{47}{8} \)

Now substitute the value of \(f\) back into the expression for \(g\):

\( g = -6 - 2(\frac{47}{8}) \)

\( g = -6 - \frac{47}{4} \)

\( g = \frac{-24 - 47}{4} \)

\( g = \frac{-71}{4} \)

Finally, substitute the value of \(g\) back into the expression for \(c\):

\( c = -1 - 2g \)

\( c = -1 - 2(\frac{-71}{4}) \)

\( c = -1 - (\frac{-71}{2}) \)

\( c = -1 + \frac{71}{2} \)

\( c = \frac{-2 + 71}{2} \)

\( c = \frac{69}{2} \)

So, we have \(g = \frac{-71}{4}\), \(f = \frac{47}{8}\), and \(c = \frac{69}{2}\).

Writing the Equation of the Circle

Substitute these values into the general equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \):

\( x^2 + y^2 + 2(\frac{-71}{4})x + 2(\frac{47}{8})y + \frac{69}{2} = 0 \)

\( x^2 + y^2 - \frac{71}{2}x + \frac{47}{4}y + \frac{69}{2} = 0 \)

To match the format of the given options, which have coefficients of 4 for \(x^2\) and \(y^2\), multiply the entire equation by 4:

\( 4(x^2 + y^2 - \frac{71}{2}x + \frac{47}{4}y + \frac{69}{2}) = 4(0) \)

\( 4x^2 + 4y^2 - 142x + 47y + 138 = 0 \)

This is the equation of the circle that passes through the points (1, 0), (0, -6), and (3, 4).

Verification with Options

Comparing the derived equation \(4x^2 + 4y^2 - 142x + 47y + 138 = 0\) with the provided options, it matches option 3 (allowing for a minor typo in the option text where '= 0' appears as '- 0').

Revision Table: Equation of Circle

Additional Information: Three Points Determine a Unique Circle

A fundamental concept in geometry is that three non-collinear points uniquely determine a circle. If the three points were collinear (lay on the same straight line), it would be impossible to draw a circle passing through all of them.

The method used above, substituting the points into the general equation \(x^2 + y^2 + 2gx + 2fy + c = 0\), relies on this principle. Each point provides a linear constraint on the coefficients \(g\), \(f\), and \(c\). With three points, we get three linear equations, which is typically sufficient to solve for the three unknowns, provided the points are not collinear.

Another method to find the equation of the circle passing through three points involves finding the perpendicular bisectors of the line segments formed by connecting the points. The intersection of any two of these perpendicular bisectors gives the center of the circle. The distance from the center to any of the three points is the radius. Once the center \((h, k)\) and radius \(r\) are known, the equation of the circle can be written in the standard form: \( (x-h)^2 + (y-k)^2 = r^2 \). This standard form can then be expanded to match the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \). The method used in the solution is generally more direct algebraically.