At what point on the circle `x^(2)+y^(2)-2x-4y+1=0` the tangent is parallel to
(1) X-axis
(2) Y-axis

Answer

To solve the problem, we need to find the points on the circle defined by the equation \( x^2 + y^2 - 2x - 4y + 1 = 0 \) where the tangent is parallel to the x-axis and y-axis.

### Step 1: Rewrite the Circle Equation
First, we can rewrite the given equation of the circle in standard form.

The equation is:
\[ x^2 + y^2 - 2x - 4y + 1 = 0 \]

We can complete the square for \(x\) and \(y\):

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

2. For \(y\):
\[ y^2 - 4y = (y-2)^2 - 4 \]

Substituting these back into the equation gives:
\[ (x-1)^2 - 1 + (y-2)^2 - 4 + 1 = 0 \]
\[ (x-1)^2 + (y-2)^2 - 4 = 0 \]
\[ (x-1)^2 + (y-2)^2 = 4 \]

This represents a circle centered at \((1, 2)\) with a radius of \(2\).

### Step 2: Find Points Where Tangent is Parallel to X-axis
A tangent to the circle is parallel to the x-axis when the slope of the tangent line is zero. The slope of the tangent line can be found using implicit differentiation.

Differentiating the circle equation implicitly:
\[ 2x + 2y \frac{dy}{dx} - 2 - 4 \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ (2y - 4) \frac{dy}{dx} = 2 - 2x \]
\[ \frac{dy}{dx} = \frac{2 - 2x}{2y - 4} \]

Setting the slope equal to zero for the tangent to be parallel to the x-axis:
\[ \frac{2 - 2x}{2y - 4} = 0 \]
This implies:
\[ 2 - 2x = 0 \quad \Rightarrow \quad x = 1 \]

Now, we substitute \(x = 1\) back into the circle equation to find \(y\):
\[ (1-1)^2 + (y-2)^2 = 4 \]
\[ 0 + (y-2)^2 = 4 \]
\[ (y-2)^2 = 4 \]
Taking the square root gives:
\[ y - 2 = 2 \quad \Rightarrow \quad y = 4 \quad \text{or} \quad y - 2 = -2 \quad \Rightarrow \quad y = 0 \]

Thus, the points where the tangent is parallel to the x-axis are:
\[ (1, 4) \quad \text{and} \quad (1, 0) \]

### Step 3: Find Points Where Tangent is Parallel to Y-axis
A tangent to the circle is parallel to the y-axis when the slope of the tangent line is undefined (infinity). This occurs when the denominator of the slope expression is zero:
\[ 2y - 4 = 0 \quad \Rightarrow \quad y = 2 \]

Now, we substitute \(y = 2\) back into the circle equation to find \(x\):
\[ (x-1)^2 + (2-2)^2 = 4 \]
\[ (x-1)^2 + 0 = 4 \]
\[ (x-1)^2 = 4 \]
Taking the square root gives:
\[ x - 1 = 2 \quad \Rightarrow \quad x = 3 \quad \text{or} \quad x - 1 = -2 \quad \Rightarrow \quad x = -1 \]

Thus, the points where the tangent is parallel to the y-axis are:
\[ (3, 2) \quad \text{and} \quad (-1, 2) \]

### Final Answer
1. Points where the tangent is parallel to the x-axis: \((1, 4)\) and \((1, 0)\)
2. Points where the tangent is parallel to the y-axis: \((3, 2)\) and \((-1, 2)\)