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

Answer

To find 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 Equation
We start with the circle equation:
\[ x^2 + y^2 + 4x - 2y - 1 = 0 \]
We can rewrite this equation in the standard form of a circle.

### Step 2: Identify the Circle's Center and Radius
To find the center and radius, 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 circle equation gives:
\[ (x + 2)^2 - 4 + (y - 1)^2 - 1 - 1 = 0 \]
Simplifying this, we have:
\[ (x + 2)^2 + (y - 1)^2 - 6 = 0 \]
\[ (x + 2)^2 + (y - 1)^2 = 6 \]

From this, we can see that the center of the circle is \((-2, 1)\) and the radius \(r\) is \(\sqrt{6}\).

### Step 3: Substitute the Point into the Circle Equation
Now, we will substitute the point (1, -2) into the left-hand side of the original circle equation to determine its position relative to the circle.

Let \(x_1 = 1\) and \(y_1 = -2\):
\[ S_1 = (1)^2 + (-2)^2 + 4(1) - 2(-2) - 1 \]
Calculating this:
\[ S_1 = 1 + 4 + 4 + 4 - 1 \]
\[ S_1 = 12 \]

### Step 4: Determine the Position of the Point
Now we evaluate \(S_1\):
- If \(S_1 < 0\), the point is inside the circle.
- If \(S_1 = 0\), the point is on the circle.
- If \(S_1 > 0\), the point is outside the circle.

Since \(S_1 = 12 > 0\), we conclude that the point (1, -2) lies **outside** the circle.

### Summary
The position of the point (1, -2) with respect to the circle \(x^2 + y^2 + 4x - 2y - 1 = 0\) is **outside** the circle.

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