Let the centre of a circle, passing through the points $$(0, 0)$$, $$(1, 0)$$ and touching the circle $$x^2 + y^2 = 9$$, be $$(h, k)$$. Then for all possible values of the coordinates of the centre $$(h, k)$$, $$4(h^2 + k^2)$$ is equal to _________

Let the required circle have centre $$(h,k)$$ and radius $$R$$.

The circle passes through $$(0,0)$$ and $$(1,0)$$, so

$$h^{2}+k^{2}=R^{2} \quad -(1)$$
$$(h-1)^{2}+k^{2}=R^{2} \quad -(2)$$

Subtract $$(1)$$ from $$(2)$$:

$$(h-1)^{2}-h^{2}=0 \;\Longrightarrow\; h^{2}-2h+1-h^{2}=0 \;\Longrightarrow\; -2h+1=0$$

Hence $$h=\frac12$$.

The given circle $$x^{2}+y^{2}=9$$ has centre $$(0,0)$$ and radius $$3$$. Distance between the two centres is therefore

$$D=\sqrt{h^{2}+k^{2}}=R \quad\text{(from }(1)\text{)}.$$

For two circles to be tangent internally, the distance between their centres equals the difference of their radii: $$|3-R|=D$$.

Substituting $$D=R$$ gives $$|3-R|=R$$, which splits into

$$3-R=R \;\Longrightarrow\; 3=2R \;\Longrightarrow\; R=\frac32$$ (possible because $$3\gt R$$),
or $$R-3=R \;\Longrightarrow\; 3=0$$ (impossible).

Thus $$R=\frac32$$, and from $$(1)$$

$$h^{2}+k^{2}=R^{2}=\left(\frac32\right)^{2}=\frac94.$$(Since $$h=\frac12$$)

$$\left(\frac12\right)^{2}+k^{2}=\frac94 \;\Longrightarrow\; \frac14+k^{2}=\frac94 \;\Longrightarrow\; k^{2}=2 \;\Longrightarrow\; k=\pm\sqrt2.$$

Finally,

$$4(h^{2}+k^{2})=4\left(\frac94\right)=9.$$

Therefore, for both possible centres $$\left(\frac12,\sqrt2\right)$$ and $$\left(\frac12,-\sqrt2\right)$$, the value of $$4(h^{2}+k^{2})$$ is $$\mathbf{9}$$.