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

Explanation

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

### Step 1: Rewrite the Circle's Equation
The equation of the circle is given as:
\[ x^2 + y^2 - 2x + 6y + 1 = 0 \]
We can rearrange this equation to the standard form of a circle. To do this, we will complete the square for both \(x\) and \(y\).
...