If the lines `2x+3y+1=0, 6x+4y+1=0` intersect the co-ordinate axes in 4 points, then the circle passing through the points is

Options

A. `12x^(2)+12y^(2)+8x+7y+1=0`
B. `6x^(2)+6y^(2)+3x+y=0`
C. `12x^(2)+12y^(2)+8x+7y+3=0`
D. `x^(2)+y^(2)+4x-y+3=0`

Answer

To find the equation of the circle that passes through the points where the lines \(2x + 3y + 1 = 0\) and \(6x + 4y + 1 = 0\) intersect the coordinate axes, we will follow these steps:

### Step 1: Find the intersection points of the lines with the axes

**For the line \(2x + 3y + 1 = 0\):**

- **X-intercept:** Set \(y = 0\):
\[ 2x + 1 = 0 \implies x = -\frac{1}{2} \]
Thus, the point is \((-1/2, 0)\).

- **Y-intercept:** Set \(x = 0\):
\[ 3y + 1 = 0 \implies y = -\frac{1}{3} \]
Thus, the point is \((0, -1/3)\).

**For the line \(6x + 4y + 1 = 0\):**

- **X-intercept:** Set \(y = 0\):
\[ 6x + 1 = 0 \implies x = -\frac{1}{6} \]
Thus, the point is \((-1/6, 0)\).

- **Y-intercept:** Set \(x = 0\):
\[ 4y + 1 = 0 \implies y = -\frac{1}{4} \]
Thus, the point is \((0, -1/4)\).

### Step 2: List the points of intersection

The four points of intersection are:
1. \((-1/2, 0)\)
2. \((0, -1/3)\)
3. \((-1/6, 0)\)
4. \((0, -1/4)\)

### Step 3: Set up the general equation of the circle

The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
We need to find the values of \(g\), \(f\), and \(c\) such that the circle passes through the four points.

### Step 4: Substitute the points into the circle equation

1. For the point \((-1/2, 0)\):
\[ \left(-\frac{1}{2}\right)^2 + 0^2 + 2g\left(-\frac{1}{2}\right) + 2f(0) + c = 0 \]
\[ \frac{1}{4} - g + c = 0 \quad \text{(Equation 1)} \]

2. For the point \((0, -1/3)\):
\[ 0^2 + \left(-\frac{1}{3}\right)^2 + 2g(0) + 2f\left(-\frac{1}{3}\right) + c = 0 \]
\[ \frac{1}{9} - \frac{2f}{3} + c = 0 \quad \text{(Equation 2)} \]

3. For the point \((-1/6, 0)\):
\[ \left(-\frac{1}{6}\right)^2 + 0^2 + 2g\left(-\frac{1}{6}\right) + 2f(0) + c = 0 \]
\[ \frac{1}{36} - \frac{g}{3} + c = 0 \quad \text{(Equation 3)} \]

4. For the point \((0, -1/4)\):
\[ 0^2 + \left(-\frac{1}{4}\right)^2 + 2g(0) + 2f\left(-\frac{1}{4}\right) + c = 0 \]
\[ \frac{1}{16} - \frac{f}{2} + c = 0 \quad \text{(Equation 4)} \]

### Step 5: Solve the equations

Now we have a system of four equations:
1. \(c - g = -\frac{1}{4}\)
2. \(c - \frac{2f}{3} = -\frac{1}{9}\)
3. \(c - \frac{g}{3} = -\frac{1}{36}\)
4. \(c - \frac{f}{2} = -\frac{1}{16}\)

We can solve these equations step by step to find \(g\), \(f\), and \(c\).

### Step 6: Substitute and simplify

1. From Equation 1, express \(c\):
\[ c = g - \frac{1}{4} \]

2. Substitute \(c\) into Equation 2:
\[ g - \frac{1}{4} - \frac{2f}{3} = -\frac{1}{9} \]
Rearranging gives:
\[ g - \frac{2f}{3} = -\frac{1}{9} + \frac{1}{4} = \frac{-4 + 9}{36} = \frac{5}{36} \]

3. Substitute \(c\) into Equation 3:
\[ g - \frac{1}{4} - \frac{g}{3} = -\frac{1}{36} \]
Rearranging gives:
\[ \frac{2g}{3} = -\frac{1}{36} + \frac{1}{4} = \frac{-1 + 9}{36} = \frac{8}{36} = \frac{2}{9} \]
Thus, \(g = \frac{2}{9} \cdot \frac{3}{2} = \frac{1}{3}\).

4. Substitute \(g\) back to find \(c\):
\[ c = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12}. \]

5. Finally, substitute \(c\) into Equation 4 to find \(f\):
\[ \frac{1}{12} - \frac{f}{2} = -\frac{1}{16}. \]
Rearranging gives:
\[ \frac{f}{2} = \frac{1}{12} + \frac{1}{16} = \frac{4 + 3}{48} = \frac{7}{48} \implies f = \frac{7}{24}. \]

### Step 7: Write the final equation of the circle

Now we have \(g = \frac{1}{3}\), \(f = \frac{7}{24}\), and \(c = \frac{1}{12}\). Substitute these values into the general circle equation:
\[ x^2 + y^2 + 2\left(\frac{1}{3}\right)x + 2\left(\frac{7}{24}\right)y + \frac{1}{12} = 0. \]
Multiplying through by \(24\) to eliminate fractions:
\[ 24x^2 + 24y^2 + 16x + 14y + 2 = 0. \]
This simplifies to:
\[ 12x^2 + 12y^2 + 8x + 7y + 1 = 0. \]

### Final Answer

The equation of the circle passing through the four points is:
\[ 12x^2 + 12y^2 + 8x + 7y + 1 = 0. \]