Equation of circle passing through (-1,-2) and concentric with the circle `x^(2)+y^(2)+3x+4y+1=0`

Options

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

Answer

To find the equation of the circle that passes through the point (-1, -2) and is concentric with the circle given by the equation \(x^2 + y^2 + 3x + 4y + 1 = 0\), we will follow these steps:

### Step 1: Identify the center of the given circle
The general form of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius. We can rewrite the given equation in standard form.

The given equation is:
\[ x^2 + y^2 + 3x + 4y + 1 = 0 \]
To find the center, we complete the square for \(x\) and \(y\).

1. For \(x\):
\[ x^2 + 3x \rightarrow (x + \frac{3}{2})^2 - \frac{9}{4} \]
2. For \(y\):
\[ y^2 + 4y \rightarrow (y + 2)^2 - 4 \]

Substituting these back, we have:
\[ (x + \frac{3}{2})^2 - \frac{9}{4} + (y + 2)^2 - 4 + 1 = 0 \]
Combining the constants:
\[ (x + \frac{3}{2})^2 + (y + 2)^2 - \frac{9}{4} - 4 + 1 = 0 \]
\[ (x + \frac{3}{2})^2 + (y + 2)^2 - \frac{9}{4} - \frac{16}{4} + \frac{4}{4} = 0 \]
\[ (x + \frac{3}{2})^2 + (y + 2)^2 - \frac{21}{4} = 0 \]

Thus, the center of the circle is \((- \frac{3}{2}, -2)\).

### Step 2: Find the radius of the new circle
Since the new circle is concentric with the given circle, it will have the same center. We need to find the radius using the point (-1, -2), which lies on the new circle.

Using the distance formula to find the radius \(R\):
\[ R^2 = \left(-1 + \frac{3}{2}\right)^2 + (-2 + 2)^2 \]
Calculating:
\[ R^2 = \left(\frac{1}{2}\right)^2 + 0^2 = \frac{1}{4} \]

### Step 3: Write the equation of the new circle
The equation of the new circle with center \((- \frac{3}{2}, -2)\) and radius \(\sqrt{\frac{1}{4}} = \frac{1}{2}\) is:
\[ \left(x + \frac{3}{2}\right)^2 + (y + 2)^2 = \frac{1}{4} \]

### Step 4: Expand the equation
Expanding this gives:
\[ \left(x + \frac{3}{2}\right)^2 + (y + 2)^2 = \frac{1}{4} \]
\[ x^2 + 3x + \frac{9}{4} + y^2 + 4y + 4 = \frac{1}{4} \]
Combining terms:
\[ x^2 + y^2 + 3x + 4y + \frac{9}{4} + 4 - \frac{1}{4} = 0 \]
\[ x^2 + y^2 + 3x + 4y + \frac{9 + 16 - 1}{4} = 0 \]
\[ x^2 + y^2 + 3x + 4y + \frac{24}{4} = 0 \]
\[ x^2 + y^2 + 3x + 4y + 6 = 0 \]

### Final Answer
Thus, the equation of the circle is:
\[ x^2 + y^2 + 3x + 4y + 6 = 0 \]