If the length of the chord of the circle, $$x^2 + y^2 = r^2$$ $$(r > 0)$$ along the line, $$y - 2x = 3$$ is $$r$$, then $$r^2$$ is equal to:

We are given the circle $$x^{2}+y^{2}=r^{2}$$, whose centre is clearly at the origin $$(0,0)$$ and whose radius is $$r$$.

The chord lies on the straight line $$y-2x=3$$. Re-write this line in the standard linear form $$Ax+By+C=0$$, because the perpendicular-distance formula uses that form:

$$y-2x-3=0\;\;\Longrightarrow\;\;-2x+y-3=0.$$

For convenience we multiply by $$-1$$ so that the coefficient of $$x$$ is positive:

$$2x-y+3=0.$$

The perpendicular distance $$d$$ from a point $$(x_{0},y_{0})$$ to the line $$Ax+By+C=0$$ is given by the formula

$$d=\frac{|Ax_{0}+By_{0}+C|}{\sqrt{A^{2}+B^{2}}}.$$

Here our point is the centre $$(0,0)$$, and the coefficients are $$A=2,\;B=-1,\;C=3$$. Substituting, we obtain

$$d=\frac{|\,2\cdot0+(-1)\cdot0+3\,|}{\sqrt{2^{2}+(-1)^{2}}} =\frac{|\,3\,|}{\sqrt{4+1}} =\frac{3}{\sqrt{5}}.$$

So,

$$d^{2}=\left(\frac{3}{\sqrt{5}}\right)^{2}=\frac{9}{5}.$$

Now recall the chord-length formula in a circle: if a chord is at a perpendicular distance $$d$$ from the centre and the radius of the circle is $$r$$, then its length $$L$$ is

$$L=2\sqrt{\,r^{2}-d^{2}\,}.$$

According to the problem, the length of this chord is exactly $$r$$ itself. Hence we set

$$r=2\sqrt{\,r^{2}-d^{2}\,}.$$

Square both sides to eliminate the square root:

$$r^{2}=4\bigl(r^{2}-d^{2}\bigr).$$

Expand the right side and gather like terms:

$$r^{2}=4r^{2}-4d^{2}\;\;\Longrightarrow\;\;4d^{2}=4r^{2}-r^{2}=3r^{2}.$$

Divide by $$3$$ to isolate $$r^{2}$$:

$$r^{2}=\frac{4}{3}d^{2}.$$

We already found $$d^{2}=\dfrac{9}{5}$$, so substitute:

$$r^{2}=\frac{4}{3}\times\frac{9}{5} =\frac{36}{15} =\frac{12}{5}.$$

Hence, the correct answer is Option D.