Let a circle $$C : (x - h)^2 + (y - k)^2 = r^2, k > 0$$, touch the $$x$$-axis at $$(1, 0)$$. If the line $$x + y = 0$$ intersects the circle $$C$$ at $$P$$ and $$Q$$ such that the length of the chord $$PQ$$ is $$2$$, then the value of $$h + k + r$$ is equal to ______.

The circle $$C: (x-h)^2 + (y-k)^2 = r^2$$ with $$k > 0$$ touches the x-axis at $$(1, 0)$$.

Since the circle touches the x-axis at $$(1, 0)$$, the center must be directly above this point, so $$h = 1$$ and $$k = r$$ (since the distance from center to x-axis equals the radius).

The circle is: $$(x-1)^2 + (y-r)^2 = r^2$$

The line $$x + y = 0$$ intersects the circle such that chord $$PQ$$ has length 2.

The distance from the center $$(1, r)$$ to the line $$x + y = 0$$ is:

$$d = \frac{|1 + r|}{\sqrt{2}} = \frac{1 + r}{\sqrt{2}}$$ (since $$r > 0$$)

Using the chord length formula: $$\left(\frac{PQ}{2}\right)^2 = r^2 - d^2$$

$$1 = r^2 - \frac{(1+r)^2}{2}$$

$$2 = 2r^2 - (1 + r)^2$$

$$2 = 2r^2 - 1 - 2r - r^2$$

$$2 = r^2 - 2r - 1$$

$$r^2 - 2r - 3 = 0$$

$$(r - 3)(r + 1) = 0$$

Since $$r > 0$$, we get $$r = 3$$.

Therefore: $$h + k + r = 1 + 3 + 3 = 7$$

The correct answer is $$7$$.