A diameter of `x^(2)+y^(2)-2x-6y +6=0` is a chord to circle centre (2, 1), then radius of the circle is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the radius of the circle centered at (2, 1) given that a diameter of the circle defined by the equation \(x^2 + y^2 - 2x - 6y + 6 = 0\) is a chord of the second circle, we can follow these steps:

### Step 1: Rewrite the equation of the first circle
The equation of the first circle is given as:
\[ x^2 + y^2 - 2x - 6y + 6 = 0 \]
We can rearrange this into the standard form of a circle.

### Step 2: Complete the square
To convert the equation into standard form, we complete the square for both \(x\) and \(y\).

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

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

Substituting these back into the equation:
\[ (x - 1)^2 - 1 + (y - 3)^2 - 9 + 6 = 0 \]
Simplifying this gives:
\[ (x - 1)^2 + (y - 3)^2 - 4 = 0 \]
Thus, we have:
\[ (x - 1)^2 + (y - 3)^2 = 4 \]

### Step 3: Identify the center and radius of the first circle
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center: \((h, k) = (1, 3)\)
- Radius: \(r = \sqrt{4} = 2\)

### Step 4: Find the distance from the center of the second circle to the center of the first circle
The center of the second circle is given as \((2, 1)\). We need to find the distance \(OC\) between the centers \((1, 3)\) and \((2, 1)\):
\[ OC = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 5: Use the Pythagorean theorem to find the radius of the second circle
Let \(OA\) be the radius of the second circle, and \(AC\) be the radius of the first circle (which we found to be 2). According to the Pythagorean theorem:
\[ OA^2 = OC^2 + AC^2 \]
Substituting the known values:
\[ OA^2 = (\sqrt{5})^2 + 2^2 \]
\[ OA^2 = 5 + 4 = 9 \]
Thus,
\[ OA = \sqrt{9} = 3 \]

### Conclusion
The radius of the second circle is \(3\).