Find the radius and centre of the circle of the circle `x^(2) + y^(2) + 2x + 4y -1=0`.

Explanation

To find the radius and center of the circle given by the equation \( x^2 + y^2 + 2x + 4y - 1 = 0 \), we can follow these steps:

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

### Step 2: Compare with the general form
The general form of a circle's equation is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From our equation, we can identify:
- \(2g = 2\) (coefficient of \(x\))
- \(2f = 4\) (coefficient of \(y\))
- \(c = -1\)

### Step 3: Solve for \(g\) and \(f\)
From the comparisons:
- \(g = 1\)
- \(f = 2\)

### Step 4: Find the center of the circle
The center of the circle \((h, k)\) can be found using:
\[ (h, k) = (-g, -f) \]
Substituting the values of \(g\) and \(f\):
\[ (h, k) = (-1, -2) \]

### Step 5: Calculate the radius
The radius \(r\) can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values of \(g\), \(f\), and \(c\):
\[ r = \sqrt{1^2 + 2^2 - (-1)} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

### Final Answer
Thus, the center of the circle is \((-1, -2)\) and the radius is \(\sqrt{6}\).

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