Find the position of the point (1,-2) with respect to the circle `x^(2)+y^(2)+4x-2y-1=0`.

Answer

To determine the position of the point (1, -2) with respect to the circle given by the equation \( x^2 + y^2 + 4x - 2y - 1 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:

\[ x^2 + y^2 + 4x - 2y - 1 = 0 \]

We can rearrange it to:

\[ x^2 + 4x + y^2 - 2y = 1 \]

### Step 2: Complete the Square
Next, we complete the square for the \(x\) and \(y\) terms.

For \(x^2 + 4x\):
\[ x^2 + 4x = (x + 2)^2 - 4 \]

For \(y^2 - 2y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]

Substituting these back into the equation gives:

\[ (x + 2)^2 - 4 + (y - 1)^2 - 1 = 1 \]

This simplifies to:

\[ (x + 2)^2 + (y - 1)^2 = 6 \]

### Step 3: Identify the Center and Radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:

- Center \((h, k) = (-2, 1)\)
- Radius \(r = \sqrt{6}\)

### Step 4: Calculate the Distance from the Point to the Center
Now we need to find the distance from the point \(M(1, -2)\) to the center of the circle \((-2, 1)\) using the distance formula:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting the coordinates:

\[ d = \sqrt{(1 - (-2))^2 + (-2 - 1)^2} \]
\[ = \sqrt{(1 + 2)^2 + (-3)^2} \]
\[ = \sqrt{3^2 + (-3)^2} \]
\[ = \sqrt{9 + 9} \]
\[ = \sqrt{18} = 3\sqrt{2} \]

### Step 5: Compare the Distance with the Radius
Now we compare the distance \(d = 3\sqrt{2}\) with the radius \(r = \sqrt{6}\).

Since \(3\sqrt{2} > \sqrt{6}\) (as \(3\sqrt{2} \approx 4.24\) and \(\sqrt{6} \approx 2.45\)), we conclude that the point lies outside the circle.

### Conclusion
The point \( (1, -2) \) lies outside the circle defined by the equation \( x^2 + y^2 + 4x - 2y - 1 = 0 \).

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