Discuss the position of the points (1,2) and (6,0) with respect to the circle `x^2+y^2-4x+2y-11=0`

Answer

To determine the position of the points (1, 2) and (6, 0) with respect to the circle defined by the equation \( x^2 + y^2 - 4x + 2y - 11 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 - 4x + 2y - 11 = 0 \]
We can rewrite this in the standard form of a circle, which is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
To do this, we will complete the square for the \(x\) and \(y\) terms.

### Step 2: Completing the Square
1. For \(x\):
\[ x^2 - 4x \rightarrow (x - 2)^2 - 4 \]
2. For \(y\):
\[ y^2 + 2y \rightarrow (y + 1)^2 - 1 \]

Now substituting back into the equation:
\[ (x - 2)^2 - 4 + (y + 1)^2 - 1 - 11 = 0 \]
This simplifies to:
\[ (x - 2)^2 + (y + 1)^2 - 16 = 0 \]
Thus, we have:
\[ (x - 2)^2 + (y + 1)^2 = 16 \]

### Step 3: Identify the Center and Radius
From the equation \((x - 2)^2 + (y + 1)^2 = 16\), we can identify:
- Center \((h, k) = (2, -1)\)
- Radius \(r = \sqrt{16} = 4\)

### Step 4: Determine the Position of Point (1, 2)
Let \(P_1(1, 2)\):
- Substitute \(x = 1\) and \(y = 2\) into the circle equation:
\[ s_1 = (1 - 2)^2 + (2 + 1)^2 - 16 \]
Calculating:
\[ s_1 = (-1)^2 + (3)^2 - 16 = 1 + 9 - 16 = -6 \]
Since \(s_1 < 0\), point \(P_1(1, 2)\) is **inside** the circle.

### Step 5: Determine the Position of Point (6, 0)
Let \(P_2(6, 0)\):
- Substitute \(x = 6\) and \(y = 0\) into the circle equation:
\[ s_2 = (6 - 2)^2 + (0 + 1)^2 - 16 \]
Calculating:
\[ s_2 = (4)^2 + (1)^2 - 16 = 16 + 1 - 16 = 1 \]
Since \(s_2 > 0\), point \(P_2(6, 0)\) is **outside** the circle.

### Final Conclusion
- The point \( (1, 2) \) is **inside** the circle.
- The point \( (6, 0) \) is **outside** the circle.

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