1. Use the given axis and distances to locate vertex and focus

Since the axis is the line $y=x$, both the vertex and focus must lie on this line.

Any point on $y=x$ in the first quadrant has form $(t,t)$. Its distance from origin is $\sqrt{t^2+t^2}=t\sqrt{2}.$

• Vertex is at distance $\sqrt{2}$ from origin, so $t\sqrt{2}=\sqrt{2}\implies t=1.$ Hence vertex is $V=(1,1).$

• Focus is at distance $2\sqrt{2}$ from origin, so $t\sqrt{2}=2\sqrt{2}\implies t=2.$ Hence focus is $F=(2,2).$

So the parabola has axis along $y=x$, vertex $(1,1)$, and opens in the direction from $(1,1)$ to $(2,2)$.

2. Find the parameter $a$

The distance from vertex to focus is $VF=\sqrt{(2-1)^2+(2-1)^2}=\sqrt{2}.$

Thus the focal length is $a=\sqrt{2}.$

3. Rotate coordinates to align with the axis

Let us use coordinates centered at the vertex and aligned with the axis.

Define $X=x-1,\qquad Y=y-1.$

Now rotate axes by $45^\circ$ so that the new $u$-axis lies along $y=x$: $u=\frac{X+Y}{\sqrt{2}},\qquad v=\frac{Y-X}{\sqrt{2}}.$

For a parabola with vertex at origin and axis along positive $u$-axis, the standard form is $v^2=4au.$

Since $a=\sqrt{2}$, $v^2=4\sqrt{2}\,u.$

Substitute $u,v$: $\left(\frac{Y-X}{\sqrt{2}}\right)^2=4\sqrt{2}\left(\frac{X+Y}{\sqrt{2}}\right).$

Simplifying, $\frac{(Y-X)^2}{2}=4(X+Y)$ $\implies (Y-X)^2=8(X+Y).$

Now put back $X=x-1,\ Y=y-1$: $\big((y-1)-(x-1)\big)^2=8\big((x-1)+(y-1)\big).$

This gives $(y-x)^2=8(x+y-2).$

So the parabola is $\boxed{(y-x)^2=8(x+y-2)}.$

4. Use the point $(1,k)$

Substitute $x=1$, $y=k$: $(k-1)^2=8(1+k-2)=8(k-1).$

So $(k-1)^2-8(k-1)=0$ $(k-1)\big((k-1)-8\big)=0$ $(k-1)(k-9)=0.$

Hence $k=1 \quad \text{or} \quad k=9.$

Among the given options, the possible value is $\boxed{9}.$

So the correct option is C.

5. Comparison with stored answer

Stored correct answer: C

Our derived answer: C (i.e. $k=9$)

They agree.