The position of point (1, 2) w.r.t. the circle `x^(2)+y^(2)-2x+6y+1=0` is

Options

A. inside
B. outside
C. on
D. none

Answer

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

### Step 1: Rewrite the equation of the circle
First, we need to 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 as:
\[ x^2 - 2x + y^2 + 6y + 1 = 0 \]

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

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:
\[ (x - 1)^2 - 1 + (y + 3)^2 - 9 + 1 = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y + 3)^2 - 9 = 0 \]
or
\[ (x - 1)^2 + (y + 3)^2 = 9 \]

### Step 3: Identify the center and radius of the circle
From the equation \((x - 1)^2 + (y + 3)^2 = 9\), we can see that the center of the circle is at \( (1, -3) \) and the radius \( r \) is \( \sqrt{9} = 3 \).

### Step 4: Calculate the distance from the point to the center of the circle
Now, we need to calculate the distance from the point \( (1, 2) \) to the center \( (1, -3) \):
\[ \text{Distance} = \sqrt{(1 - 1)^2 + (2 - (-3))^2} = \sqrt{0 + (2 + 3)^2} = \sqrt{5^2} = 5 \]

### Step 5: Compare the distance with the radius
Now we compare the distance from the point to the center of the circle (which is 5) with the radius (which is 3):
- If the distance is greater than the radius, the point is outside the circle.
- If the distance is equal to the radius, the point is on the circle.
- If the distance is less than the radius, the point is inside the circle.

In this case, since \( 5 > 3 \), the point \( (1, 2) \) is outside the circle.

### Final Answer
Thus, the position of the point \( (1, 2) \) with respect to the circle is **outside** the circle.

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