The radius of the circle, having centre at (2, 1) whose one of the chord is a diameter of the circle `x^(2)+y^(2)-2x -6y +6=0` is

Options

A. 1
B. 2
C. 3
D. `sqrt(3)`

Answer

To find the radius of the circle with center at (2, 1) whose one chord is a diameter of the circle given by the equation \(x^2 + y^2 - 2x - 6y + 6 = 0\), we will follow these steps:

### Step 1: Find the center and radius of the given circle

The equation of the circle is given as:
\[ x^2 + y^2 - 2x - 6y + 6 = 0 \]
We can rewrite this equation in standard form by completing the square.

1. Rearranging the equation:
\[ (x^2 - 2x) + (y^2 - 6y) + 6 = 0 \]

2. Completing the square for \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

3. Completing the square for \(y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

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

From this, we can see that the center of the circle \(C_1\) is \((1, 3)\) and the radius \(r_1\) is:
\[ r_1 = \sqrt{4} = 2 \]

### Step 2: Find the diameter of the circle

The diameter \(d_1\) of the circle is twice the radius:
\[ d_1 = 2 \times r_1 = 2 \times 2 = 4 \]

### Step 3: Calculate the distance between the centers of the two circles

We have the center of the new circle \(C_2\) at \((2, 1)\) and the center of the given circle \(C_1\) at \((1, 3)\). We will find the distance \(C_1C_2\) using the distance formula:
\[ C_1C_2 = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 4: Use Pythagorean theorem to find the radius of the larger circle

Let \(A\) be the midpoint of the chord (which is also the diameter of the given circle). We know:
- \(C_1C_2 = \sqrt{5}\)
- The radius of the circle \(C_1\) is \(r_1 = 2\)

Using the Pythagorean theorem in triangle \(AC_1C_2\):
\[ AC_2^2 = AC_1^2 + C_1C_2^2 \]
Let \(AC_2\) be the radius \(R\) of the larger circle. Then:
\[ R^2 = 2^2 + (\sqrt{5})^2 \]
\[ R^2 = 4 + 5 = 9 \]
\[ R = \sqrt{9} = 3 \]

### Conclusion

The radius of the circle with center at (2, 1) whose one chord is a diameter of the given circle is:
\[ \boxed{3} \]