The radius of the circle with centre at `(3, 2)` and whose common chord with the cirlce `C:x^(2)+y^(2)-4x-8y+16=0` is also a diameter of the circle C, is

Options

A. 3 units
B. 2 units
C. 1 units
D. `sqrt3` units

Answer

To find the radius of the circle with center at (3, 2) and whose common chord with the circle \( C: x^2 + y^2 - 4x - 8y + 16 = 0 \) is also a diameter of circle \( C \), we will follow these steps:

### Step 1: Find the center and radius of circle C
The equation of circle \( C \) is given by:
\[ x^2 + y^2 - 4x - 8y + 16 = 0 \]
We can rewrite this in standard form by completing the square.

1. Rearranging the equation:
\[ (x^2 - 4x) + (y^2 - 8y) + 16 = 0 \]

2. Completing the square:
- For \( x^2 - 4x \):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
- For \( y^2 - 8y \):
\[ y^2 - 8y = (y - 4)^2 - 16 \]

3. Substitute back into the equation:
\[ (x - 2)^2 - 4 + (y - 4)^2 - 16 + 16 = 0 \]
Simplifying gives:
\[ (x - 2)^2 + (y - 4)^2 - 4 = 0 \]
Thus,
\[ (x - 2)^2 + (y - 4)^2 = 4 \]

From this, we can see that the center of circle \( C \) is \( (2, 4) \) and the radius \( r_C = \sqrt{4} = 2 \).

### Step 2: Identify the midpoint of the common chord
Since the common chord is also the diameter of circle \( C \), the midpoint of this chord (which is also the center of circle \( C \)) is \( M(2, 4) \).

### Step 3: Calculate the distance from the center of the required circle to the midpoint
The center of the required circle is given as \( O(3, 2) \). We need to find the distance \( OM \) between points \( O \) and \( M \).

Using the distance formula:
\[ OM = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting \( O(3, 2) \) and \( M(2, 4) \):
\[ OM = \sqrt{(3 - 2)^2 + (2 - 4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 4: Calculate the radius of the required circle
Let \( r \) be the radius of the required circle. The radius can be calculated using the Pythagorean theorem, where \( AM \) is the radius of circle \( C \) and \( OM \) is the distance we just calculated:
\[ r^2 = AM^2 + OM^2 \]
Substituting \( AM = 2 \) and \( OM = \sqrt{5} \):
\[ r^2 = 2^2 + (\sqrt{5})^2 = 4 + 5 = 9 \]
Thus,
\[ r = \sqrt{9} = 3 \]

### Final Answer
The radius of the required circle is \( \boxed{3} \).