The point where the line x=0 touches the circle `x^2+y^2-2x-6y+9=0` is

Options

A. (0,1)
B. (0,2)
C. (0,3)
D. no where

Answer

To find the point where the line \( x = 0 \) touches the circle given by the equation \( x^2 + y^2 - 2x - 6y + 9 = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the circle equation in standard form. The given equation is:
\[ x^2 + y^2 - 2x - 6y + 9 = 0 \]
We can rearrange it as:
\[ x^2 - 2x + y^2 - 6y + 9 = 0 \]

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

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

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

Substituting these back into the equation gives:
\[ (x - 1)^2 - 1 + (y - 3)^2 - 9 + 9 = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y - 3)^2 - 1 = 0 \]
or
\[ (x - 1)^2 + (y - 3)^2 = 1 \]

### Step 3: Identify the Center and Radius
From the standard form \( (x - h)^2 + (y - k)^2 = r^2 \), we can identify:
- Center: \( (1, 3) \)
- Radius: \( r = 1 \)

### Step 4: Find the Intersection with the Line \( x = 0 \)
To find the points where the circle intersects the line \( x = 0 \), we substitute \( x = 0 \) into the circle's equation:
\[ (0 - 1)^2 + (y - 3)^2 = 1 \]
This simplifies to:
\[ 1 + (y - 3)^2 = 1 \]
Subtracting 1 from both sides gives:
\[ (y - 3)^2 = 0 \]
Taking the square root of both sides, we find:
\[ y - 3 = 0 \implies y = 3 \]

### Step 5: Write the Point of Tangency
Thus, the point where the line \( x = 0 \) touches the circle is:
\[ (0, 3) \]

### Final Answer
The point where the line \( x = 0 \) touches the circle is \( (0, 3) \).

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