The line $$x = y$$ touches a circle at the point (1, 1). If the circle also passes through the point (1, -3), then its radius is

We have a circle which is tangent to the line $$x = y$$ at the point $$(1,\,1)$$ and which also passes through the point $$(1,\,-3)$$. Our aim is to determine the radius of this circle.

Because a radius drawn to the point of contact is always perpendicular to the tangent line, the centre of the circle must lie on the straight line that is perpendicular to $$x = y$$ at the point $$(1,\,1)$$.

The given tangent $$x = y$$ has slope $$1$$. A line perpendicular to it therefore has slope $$-1$$. Writing the equation of the line with slope $$-1$$ through $$(1,\,1)$$, we start from the point-slope form $$y - y_1 = m\,(x - x_1)$$. Substituting $$m=-1$$ and $$(x_1,\,y_1)=(1,\,1)$$, we get

$$y - 1 = -1\,(x - 1).$$

Simplifying,

$$y - 1 = -x + 1,$$

$$x + y = 2.$$

Hence the centre $$C(h,\,k)$$ of the circle must satisfy the linear relation

$$h + k = 2. \quad -(1)$$

Next, we use the distance formula. The radius $$r$$ is the distance from the centre to any point on the circle. First, taking the point of contact $$(1,\,1)$$, we have

$$r^2 = (h - 1)^2 + (k - 1)^2. \quad -(2)$$

The circle also passes through the point $$(1,\,-3)$$, so the distance from the centre to this point must equal the same radius. Therefore,

$$r^2 = (h - 1)^2 + (k + 3)^2. \quad -(3)$$

Because both right-hand sides equal $$r^2$$, we can equate them directly. Substituting from (2) and (3),

$$(h - 1)^2 + (k + 3)^2 = (h - 1)^2 + (k - 1)^2.$$

The term $$(h - 1)^2$$ appears on both sides, so it cancels, leaving

$$(k + 3)^2 = (k - 1)^2.$$

Now we expand both squares step by step:

$$k^2 + 6k + 9 = k^2 - 2k + 1.$$

Subtracting $$k^2$$ from each side eliminates the quadratic terms:

$$6k + 9 = -2k + 1.$$

Adding $$2k$$ to both sides gathers the $$k$$ terms together:

$$8k + 9 = 1.$$

Subtracting $$9$$ from both sides,

$$8k = -8.$$

Dividing by $$8$$, we find

$$k = -1.$$

Using relation (1), $$h + k = 2$$, we substitute $$k = -1$$ to get

$$h + (-1) = 2,$$

$$h = 3.$$

Thus the centre of the circle is $$C(3,\,-1).$$

Finally, we compute the radius using the distance from the centre to $$(1,\,1)$$. By the distance formula,

$$r = \sqrt{(3 - 1)^2 + (-1 - 1)^2}.$$

Calculating the squares explicitly,

$$r = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}.$$

Since $$\sqrt{8} = 2\sqrt{2},$$ we have

$$r = 2\sqrt{2}.$$

Hence, the correct answer is Option D.