Let C be the circle with centre (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

Options

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

Answer

To solve the problem, we need to find the radius of circle T, which is centered at (0, y), passes through the origin, 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 properties of circle C:**
- Center: \( C(1, 1) \)
- Radius: \( r_C = 1 \)

2. **Identify the properties of circle T:**
- Center: \( T(0, y) \)
- Radius: \( r_T = y \) (since it passes through the origin, the radius is equal to the y-coordinate of the center)

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

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

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

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

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

8. **Simplify the equation:**
- Cancel \( y^2 \) from both sides:
\[ -2y + 2 = 2y + 1 \]
- Rearranging gives:
\[ 2 - 1 = 2y + 2y \]
\[ 1 = 4y \]

9. **Solve for \( y \):**
\[ y = \frac{1}{4} \]

10. **Conclusion:**
- The radius of circle T is \( r_T = y = \frac{1}{4} \).

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