The equation of the circle which passes through the points (1,0) , (0,-6) and (3,4) is

Options

A. `4x^(2) + 4y^(2) + 142x + 47y + 140 = 0`
B. `4x^(2) + 4y^(2) - 142x - 47y + 138 =0`
C. `4x^(2) + 4y^(2) - 142x + 47y + 138 =0`
D. `4x^(2) + 4y^(2) + 150 x -49y + 138 = 0`

Answer

To find the equation of the circle that passes through the points (1, 0), (0, -6), and (3, 4), we can use the general equation of a circle:

\[ x^2 + y^2 + Dx + Ey + F = 0 \]

We need to find the coefficients \(D\), \(E\), and \(F\) such that the circle passes through the given points.

### Step 1: Substitute the first point (1, 0)

Substituting the point (1, 0) into the equation:

\[ 1^2 + 0^2 + D(1) + E(0) + F = 0 \]

This simplifies to:

\[ 1 + D + F = 0 \quad \text{(Equation 1)} \]

### Step 2: Substitute the second point (0, -6)

Now, substituting the point (0, -6):

\[ 0^2 + (-6)^2 + D(0) + E(-6) + F = 0 \]

This simplifies to:

\[ 36 - 6E + F = 0 \quad \text{(Equation 2)} \]

### Step 3: Substitute the third point (3, 4)

Next, substituting the point (3, 4):

\[ 3^2 + 4^2 + D(3) + E(4) + F = 0 \]

This simplifies to:

\[ 9 + 16 + 3D + 4E + F = 0 \]

Combining terms gives us:

\[ 25 + 3D + 4E + F = 0 \quad \text{(Equation 3)} \]

### Step 4: Solve the system of equations

Now we have a system of three equations:

1. \(D + F = -1\) (from Equation 1)
2. \(-6E + F = -36\) (from Equation 2)
3. \(3D + 4E + F = -25\) (from Equation 3)

We can express \(F\) from Equation 1:

\[ F = -1 - D \]

Substituting \(F\) into Equation 2:

\[ -6E - 1 - D = -36 \]

This simplifies to:

\[ -6E - D = -35 \quad \text{(Equation 4)} \]

Now substituting \(F\) into Equation 3:

\[ 3D + 4E - 1 - D = -25 \]

This simplifies to:

\[ 2D + 4E = -24 \quad \text{(Equation 5)} \]

### Step 5: Solve Equations 4 and 5

Now we can solve Equations 4 and 5:

From Equation 4:

\[ D = -35 + 6E \]

Substituting \(D\) into Equation 5:

\[ 2(-35 + 6E) + 4E = -24 \]

This simplifies to:

\[ -70 + 12E + 4E = -24 \]

Combining terms gives:

\[ 16E = 46 \implies E = \frac{46}{16} = \frac{23}{8} \]

### Step 6: Find D using E

Substituting \(E\) back into Equation 4 to find \(D\):

\[ D = -35 + 6 \cdot \frac{23}{8} \]

Calculating gives:

\[ D = -35 + \frac{138}{8} = -35 + 17.25 = -17.75 = -\frac{71}{4} \]

### Step 7: Find F using D

Now substituting \(D\) back into Equation 1 to find \(F\):

\[ F = -1 - \left(-\frac{71}{4}\right) = -1 + \frac{71}{4} = \frac{71}{4} - \frac{4}{4} = \frac{67}{4} \]

### Step 8: Write the equation of the circle

Now we have:

- \(D = -\frac{71}{4}\)
- \(E = \frac{23}{8}\)
- \(F = \frac{67}{4}\)

Substituting these values into the general equation of the circle:

\[ x^2 + y^2 - \frac{71}{4}x + \frac{23}{8}y + \frac{67}{4} = 0 \]

To eliminate fractions, multiply through by 8:

\[ 8x^2 + 8y^2 - 142x + 23y + 134 = 0 \]

Thus, the equation of the circle is:

\[ 8x^2 + 8y^2 - 142x + 23y + 134 = 0 \]