Let $$C$$ be the circle with center at $$(1, 1)$$ and radius = 1. If $$T$$ is the circle centered at $$(0, y)$$, passing through the origin and touching the circle $$C$$ externally, then the radius of $$T$$ is equal to:

We have the first circle $$C$$ with centre $$(1,1)$$ and radius $$1$$, so every point $$P(x,y)$$ on it satisfies

$$ (x-1)^2+(y-1)^2 = 1^2. $$

Let the second circle $$T$$ have centre $$(0,y)$$ and let its radius be $$r$$. Since the centre is at $$(0,y)$$, the distance from this centre to the origin $$(0,0)$$ is simply

$$ r = \sqrt{(0-0)^2+(y-0)^2}=|y|. $$

Thus the circle $$T$$ passes through the origin if and only if its radius equals $$|y|$$.

Next, the phrase “touching the circle $$C$$ externally’’ means that the distance between the two centres equals the sum of their radii. Writing this condition explicitly, we get

$$ \text{distance between }(0,y)\text{ and }(1,1)=1+r. $$

Using the distance formula first, the left‐hand side is

$$ \sqrt{(1-0)^2+(1-y)^2}=\sqrt{1+(1-y)^2}. $$

Therefore the external‐tangency condition becomes

$$ \sqrt{1+(1-y)^2}=1+|y|. $$

To remove the square root we square both sides, remembering to preserve equality:

$$ 1+(1-y)^2=(1+|y|)^2. $$

First expand the square on the left:

$$ (1-y)^2=1-2y+y^2, $$

so

$$ 1+(1-y)^2 = 1+\bigl(1-2y+y^2\bigr)=2-2y+y^2. $$

Now expand the square on the right:

$$ (1+|y|)^2 = 1 + 2|y| + |y|^2 = 1 + 2|y| + y^2. $$

(Because $$|y|^2=y^2$$ for every real $$y$$.) Substituting both expansions back into the equation we obtain

$$ 2-2y+y^2 = 1 + 2|y| + y^2. $$

The term $$y^2$$ cancels from both sides, leaving

$$ 2-2y = 1 + 2|y|. $$

Rearrange by subtracting $$1$$ from each side:

$$ 1-2y = 2|y|. $$

At this point we must consider the sign of $$y$$ because of the absolute value.

Case 1:  $$y\ge 0.$$ Then $$|y|=y$$, and the equation becomes

$$ 1-2y = 2y. $$

Adding $$2y$$ to both sides gives

$$ 1 = 4y \quad\Longrightarrow\quad y = \frac14. $$

Since $$y\ge 0$$ is satisfied, this is an acceptable solution. The radius in this case is

$$ r = |y| = \frac14. $$

Case 2:  $$y<0.$$ Then $$|y|=-y$$, and the equation becomes

$$ 1-2y = 2(-y)\;=\;-2y. $$

The left‐hand side is still $$1-2y$$, so equating gives

$$ 1-2y = -2y \quad\Longrightarrow\quad 1 = 0, $$

which is impossible. Therefore there is no solution with $$y<0$$.

Hence the only valid centre for the circle $$T$$ is $$(0,\tfrac14)$$, and its radius is

$$ r = \frac14. $$

Thus the radius of the required circle $$T$$ equals $$\dfrac14$$.

Hence, the correct answer is Option B.