All the points in the set $$S = \left\{\frac{\alpha + i}{\alpha - i}, \alpha \in R\right\}$$, $$i = \sqrt{-1}$$ lie on a:

Let us denote the typical element of the set by $$z$$, so we write

$$z=\frac{\alpha+i}{\alpha-i},\qquad\alpha\in\mathbb R,\qquad i=\sqrt{-1}.$$

Because we finally want the locus of $$z$$ in the Argand plane, we set

$$z=x+iy,\qquad x,y\in\mathbb R.$$

We first clear the fraction by cross-multiplication:

$$z(\alpha-i)=\alpha+i.$$

Expanding the left side, we have

$$\alpha z-iz=\alpha+i.$$

Now we isolate $$\alpha$$ because in the end we shall impose “$$\alpha$$ is real” as a condition:

$$\alpha z-\alpha=i+iz$$

$$\alpha(z-1)=i(1+z).$$

Therefore

$$\alpha=\frac{i\,(1+z)}{\,z-1\,}.$$

Since $$\alpha$$ is real, the complex quotient on the right must itself be a real number. A standard test for a complex number to be real is that its imaginary part must vanish. To evaluate that imaginary part we substitute $$z=x+iy$$:

$$1+z=(1+x)+iy,$$

$$z-1=(x-1)+iy.$$

Next we multiply the numerator by $$i$$:

$$i(1+z)=i[(1+x)+iy]=i(1+x)+i(iy)=i(1+x)-y.$$

Thus, in rectangular form,

$$i(1+z)=-y+i(1+x).$$

Putting the pieces together, the expression for $$\alpha$$ becomes

$$\alpha=\frac{-y+i(1+x)}{(x-1)+iy}.$$

To test whether this quotient is real, we multiply numerator and denominator by the conjugate of the denominator; the real denominator then does not affect the “imaginary-part = 0” criterion. Hence we examine

$$\bigl[-y+i(1+x)\bigr]\bigl[(x-1)-iy\bigr].$$

We expand this product term by term:

First part: $$(-y)(x-1)=-y(x-1).$$

Second part: $$(-y)(-iy)=i\,y^{2}.$$

Third part: $$i(1+x)(x-1)=i(1+x)(x-1).$$

Fourth part: $$i(1+x)(-iy)=-(1+x)y(-1)=+(1+x)y.$$

Collecting real and imaginary components we get

Real part: $$-y(x-1)+(1+x)y=(1+x)y-y(x-1)=y[(1+x)-(x-1)]=y[1+x-x+1]=2y.$$

Imaginary part: $$i\bigl[y^{2}+(1+x)(x-1)\bigr].$$

For $$\alpha$$ to be real, the imaginary part must be zero, so we impose

$$y^{2}+(1+x)(x-1)=0.$$

We now expand the product inside the brackets:

$$(1+x)(x-1)=x^{2}-1.$$

Substituting, the condition becomes

$$y^{2}+x^{2}-1=0,$$

or, after re-arranging,

$$x^{2}+y^{2}=1.$$

This is the standard equation of a circle centred at the origin with radius $$1$$.

Hence, the points $$z=\dfrac{\alpha+i}{\alpha-i}$$ with real $$\alpha$$ all lie on the circle $$x^{2}+y^{2}=1$$, whose radius equals $$1$$.

Hence, the correct answer is Option C.