The point (1,2) lies inside the circle `x^(2)+y^(2)-2x+6y+1=0`

Answer

To determine whether the point (1, 2) lies inside the circle given by the equation \(x^2 + y^2 - 2x + 6y + 1 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we can rewrite the equation of the circle in a more standard form. The given equation is:
\[ x^2 + y^2 - 2x + 6y + 1 = 0 \]
We can rearrange it to:
\[ x^2 - 2x + y^2 + 6y + 1 = 0 \]

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

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

For \(y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

Substituting these back into the equation gives:
\[ (x - 1)^2 - 1 + (y + 3)^2 - 9 + 1 = 0 \]
Simplifying this:
\[ (x - 1)^2 + (y + 3)^2 - 9 = 0 \]
Thus, we can rewrite it as:
\[ (x - 1)^2 + (y + 3)^2 = 9 \]

### Step 3: Identify the Circle's Center and Radius
From the equation \((x - 1)^2 + (y + 3)^2 = 9\), we can identify:
- The center of the circle is at \((1, -3)\).
- The radius of the circle is \(3\) (since \(9 = 3^2\)).

### Step 4: Calculate the Distance from the Point to the Center
Now, we need to find the distance from the point \((1, 2)\) to the center \((1, -3)\). The distance \(d\) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting in our points:
\[ d = \sqrt{(1 - 1)^2 + (2 - (-3))^2} = \sqrt{0 + (2 + 3)^2} = \sqrt{5^2} = 5 \]

### Step 5: Compare the Distance with the Radius
Now we compare the distance \(d\) with the radius \(r\):
- The distance \(d = 5\)
- The radius \(r = 3\)

Since \(d > r\), the point \((1, 2)\) lies outside the circle.

### Conclusion
Thus, the statement that the point \((1, 2)\) lies inside the circle is false.

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