How are the following points (0, 1), (3, 1) and (1,3) situated w.r.t. the circle `x^(2)+y^(2)-2x-4y+3=0`?

Answer

To determine the position of the points (0, 1), (3, 1), and (1, 3) with respect to the circle given by the equation \(x^2 + y^2 - 2x - 4y + 3 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we rewrite the circle equation in standard form by completing the square.

The given equation is:
\[ x^2 - 2x + y^2 - 4y + 3 = 0 \]

Completing the square for \(x\) and \(y\):
- For \(x^2 - 2x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
- For \(y^2 - 4y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Substituting these back into the equation:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 + 3 = 0 \]
\[ (x - 1)^2 + (y - 2)^2 - 2 = 0 \]
\[ (x - 1)^2 + (y - 2)^2 = 2 \]

This represents a circle with center \((1, 2)\) and radius \(\sqrt{2}\).

### Step 2: Evaluate Each Point
Now, we will evaluate each point by substituting their coordinates into the circle equation.

#### Point (0, 1)
Substituting \(x = 0\) and \(y = 1\):
\[ (0 - 1)^2 + (1 - 2)^2 = 1 + 1 = 2 \]
Since the result is equal to the radius squared (2), the point (0, 1) is on the circle.

#### Point (3, 1)
Substituting \(x = 3\) and \(y = 1\):
\[ (3 - 1)^2 + (1 - 2)^2 = 4 + 1 = 5 \]
Since 5 is greater than 2, the point (3, 1) is outside the circle.

#### Point (1, 3)
Substituting \(x = 1\) and \(y = 3\):
\[ (1 - 1)^2 + (3 - 2)^2 = 0 + 1 = 1 \]
Since 1 is less than 2, the point (1, 3) is inside the circle.

### Summary of Results
- The point (0, 1) is **on the circle**.
- The point (3, 1) is **outside the circle**.
- The point (1, 3) is **inside the circle**.