The parabola has its vertex at the origin and axis along the positive $$x$$-axis, so its standard form is $$y^{2}=4ax$$.
Because the point $$(4,\,4\sqrt2)$$ lies on the parabola, substitute to find $$a$$:
$$\bigl(4\sqrt2\bigr)^2=4a(4)\;\;\Longrightarrow\;\;32=16a\;\;\Longrightarrow\;\;a=2.$$
Hence the parabola is
$$y^{2}=8x\quad -(1)$$
Let the required circle be $$x^{2}+y^{2}=R^{2} \; -(2).$$
(The centre is also at the origin; we denote the radius by $$R$$.)
Using the parametric coordinates of the parabola
$$x=2t^{2},\qquad y=4t$$
(where $$t$$ is any real parameter) we impose condition $$-(2)$$:
$$\bigl(2t^{2}\bigr)^{2}+\bigl(4t\bigr)^{2}=R^{2}
\;\;\Longrightarrow\;\;4t^{4}+16t^{2}=R^{2}
\;\;\Longrightarrow\;\;R^{2}=4t^{2}\bigl(t^{2}+4\bigr). \quad -(3)$$
If the circle meets the parabola at points $$P(t_1)$$ and $$Q(t_2)$$, then both parameters $$t_1,t_2$$ satisfy $$-(3)$$ with the same $$R$$.
The chord $$PQ$$ subtends a right angle at the vertex, i.e. the origin.
So the vectors $$\overrightarrow{OP}$$ and $$\overrightarrow{OQ}$$ are perpendicular:
$$\overrightarrow{OP}\cdot\overrightarrow{OQ}=0.$$
Now
$$\overrightarrow{OP}=\bigl(2t_1^{2},\,4t_1\bigr),\qquad
\overrightarrow{OQ}=\bigl(2t_2^{2},\,4t_2\bigr).$$
Taking their dot product,
$$\bigl(2t_1^{2}\bigr)\bigl(2t_2^{2}\bigr)+\bigl(4t_1\bigr)\bigl(4t_2\bigr)=4t_1^{2}t_2^{2}+16t_1t_2
=4t_1t_2\bigl(t_1t_2+4\bigr)=0.$$
Since neither point coincides with the origin, $$t_1t_2\neq0$$, hence
$$t_1t_2+4=0\;\;\Longrightarrow\;\;t_1t_2=-4. \quad -(4)$$
Using $$-(3)$$ for both parameters:
$$R^{2}=4t_1^{2}(t_1^{2}+4)=4t_2^{2}(t_2^{2}+4).$$
Replace $$t_2$$ by $$-4/t_1$$ from $$-(4)$$ and equate the two expressions:
$$4t_1^{2}(t_1^{2}+4)=4\!\left(\frac{16}{t_1^{2}}\right)\!\left(\frac{16}{t_1^{2}}+4\right).$$
Simplifying (let $$u=t_1^{2}\;(\,u\gt0\,)$$):
$$u^{4}+4u^{3}-64u-256=0.$$
This polynomial factors as $$(u-4)(u+4)\bigl(u^{2}+4u+16\bigr)=0,$$
yielding the only positive root $$u=4.$$
Therefore $$t_1^{2}=4\;\;\Longrightarrow\;\;t_1=\pm2,$$ and by $$-(4)$$
$$t_2=-\frac{4}{t_1}=\mp2.$$
Thus the two parameters are $$t_1=2,\;t_2=-2$$ (or vice-versa).
Finally, substitute $$t_1=2$$ (or $$t_2=-2$$) into $$-(3)$$ to get the radius:
$$R^{2}=4(2)^{2}\bigl((2)^{2}+4\bigr)=4\cdot4\cdot8=128
\;\;\Longrightarrow\;\;R=\sqrt{128}=8\sqrt2.$$
Hence the radius of the circle is $$8\sqrt2$$.
Option C which is: $$8\sqrt2$$
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