Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to (1) `(sqrt(3))/(sqrt(2))` (2) `(sqrt(3))/2` (3) `1/2` (3) `1/4`

Options

A. `(sqrt(3))/(sqrt(2))`
B. `(sqrt(3))/(2)`
C. `(1)/(2)`
D. `(1)/(4)`

Explanation

To solve the problem, we need to find the radius of circle T, which is centered at (0, y) and touches circle C externally. Circle C has its center at (1, 1) and a radius of 1.

### Step-by-Step Solution:

1. **Identify the centers and radii of the circles:**
- Circle C has center \( C(1, 1) \) and radius \( r_C = 1 \).
- Circle T has center \( T(0, y) \) and radius \( r_T \).

2. **Determine the distance between the centers of the circles:**
- The distance \( d \) between the centers of circles C and T can be calculated using the distance formula:
\[ d = \sqrt{(1 - 0)^2 + (1 - y)^2} = \sqrt{1 + (1 - y)^2} \]

3. **Set up the equation for external tangency:**
- For circle T to touch circle C externally, the distance between their centers must equal the sum of their radii:
\[ d = r_C + r_T \]
- Substituting the known values:
\[ \sqrt{1 + (1 - y)^2} = 1 + r_T \]

4. **Square both sides to eliminate the square root:**
\[ 1 + (1 - y)^2 = (1 + r_T)^2 \]

5. **Expand both sides:**
- Left side:
\[ 1 + (1 - 2y + y^2) = 2 - 2y + y^2 \]
- Right side:
\[ 1 + 2r_T + r_T^2 \]

6. **Set the expanded equations equal to each other:**
\[ 2 - 2y + y^2 = 1 + 2r_T + r_T^2 \]

7. **Rearrange the equation:**
\[ y^2 - 2y + 1 = 2r_T + r_T^2 \]
\[ (y - 1)^2 = r_T^2 + 2r_T \]

8. **Recognize that \( r_T \) is the radius of circle T and substitute \( r_T = y \):**
\[ (y - 1)^2 = y^2 + 2y \]

9. **Simplify the equation:**
\[ y^2 - 2y + 1 = y^2 + 2y \]
\[ -2y + 1 = 2y \]
\[ 1 = 4y \]
\[ y = \frac{1}{4} \]

10. **Conclude the radius of circle T:**
- Since \( r_T = y \), the radius of circle T is:
\[ r_T = \frac{1}{4} \]

### Final Answer:
The radius of circle T is \( \frac{1}{4} \).