If $$\text{Re}\left(\frac{z-1}{2z+i}\right) = 1$$, where $$z = x + iy$$, then the point $$(x, y)$$ lies on a

We start by writing the complex number in the usual Cartesian form $$z = x + iy$$, where $$x$$ and $$y$$ are real numbers. The given condition is

$$\text{Re}\!\left(\dfrac{z-1}{2z+i}\right)=1.$$

First we express the numerator and denominator separately in terms of $$x$$ and $$y$$.

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

Denominator: $$2z+i=2(x+iy)+i=2x+2iy+i=2x+i(2y+1).$$

It is convenient to denote

$$a=2x,\qquad b=2y+1,$$

so that the denominator becomes $$a+ib$$. To extract the real part of a quotient we multiply the numerator and denominator by the complex conjugate of the denominator. Using the identity

$$\dfrac{p+iq}{r+is}=\dfrac{(p+iq)(r-is)}{(r+is)(r-is)}=\dfrac{(pr+qs)+i(qr-ps)}{r^{2}+s^{2}},$$

we have

$$\dfrac{(x-1)+iy}{a+ib}\;=\;\dfrac{\bigl[(x-1)+iy\bigr](a-ib)}{a^{2}+b^{2}}.$$

Carrying out the multiplication in the numerator:

$$\bigl[(x-1)+iy\bigr](a-ib)= (x-1)a -i b(x-1) + i y a - i^{2}yb.$$

Remembering that $$i^{2}=-1$$, the term $$-i^{2}yb$$ becomes $$+yb$$. So we collect real and imaginary parts:

Real part: $$(x-1)a + yb,$$

Imaginary part: $$\Bigl[-b(x-1)+ya\Bigr]i.$$

Hence

$$\dfrac{z-1}{2z+i}= \dfrac{(x-1)a+yb}{a^{2}+b^{2}}\;+\;i\,\dfrac{-b(x-1)+ya}{a^{2}+b^{2}}.$$

The real part of this fraction is therefore

$$\text{Re}\!\left(\dfrac{z-1}{2z+i}\right)=\dfrac{(x-1)a+yb}{a^{2}+b^{2}}.$$

The given condition tells us that this real part equals $$1$$, so

$$\dfrac{(x-1)a+yb}{a^{2}+b^{2}} = 1.$$

Cross-multiplying gives

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

Now we substitute back $$a=2x$$ and $$b=2y+1$$.

Left side:

$$ (x-1)(2x) + y(2y+1) = 2x(x-1) + y(2y+1). $$

Right side:

$$ a^{2}+b^{2} = (2x)^{2} + (2y+1)^{2} = 4x^{2} + (2y+1)^{2}. $$

So the equation becomes

$$2x(x-1) + y(2y+1) = 4x^{2} + (2y+1)^{2}.$$

We now expand every term.

Left side:

$$2x(x-1)=2x^{2}-2x,\qquad y(2y+1)=2y^{2}+y,$$

so

$$\text{Left side}=2x^{2}-2x+2y^{2}+y.$$

Right side:

$$4x^{2} + (2y+1)^{2}=4x^{2} + 4y^{2}+4y+1.$$

Setting left equal to right:

$$2x^{2}-2x+2y^{2}+y = 4x^{2}+4y^{2}+4y+1.$$

We now bring every term to one side to obtain zero on the other:

$$0 = 4x^{2}+4y^{2}+4y+1 - (2x^{2}-2x+2y^{2}+y).$$

Simplifying term by term:

$$0 = (4x^{2}-2x^{2}) + (4y^{2}-2y^{2}) + (4y-y) + (1-(-2x)),$$

$$0 = 2x^{2} + 2y^{2} + 3y + 2x + 1.$$

Dividing by $$2$$ to make the coefficients smaller,

$$x^{2} + y^{2} + x + \dfrac{3}{2}y + \dfrac{1}{2}=0.$$

This is recognisably the general equation of a circle. To see its centre and radius we complete the square in both $$x$$ and $$y$$.

For the $$x$$-terms:

$$x^{2}+x = x^{2}+x+\dfrac14 -\dfrac14 = \left(x+\dfrac12\right)^{2}-\dfrac14.$$

For the $$y$$-terms:

$$y^{2}+\dfrac{3}{2}y = y^{2}+\dfrac{3}{2}y+\dfrac{9}{16}-\dfrac{9}{16}= \left(y+\dfrac34\right)^{2}-\dfrac{9}{16}.$$

Substituting these back we have

$$\left(x+\dfrac12\right)^{2}-\dfrac14 \;+\; \left(y+\dfrac34\right)^{2}-\dfrac{9}{16} \;+\; \dfrac12 = 0.$$

The constants combine as

$$-\dfrac14 -\dfrac{9}{16} + \dfrac12 = -\dfrac{4}{16}-\dfrac{9}{16}+\dfrac{8}{16}=-\dfrac{5}{16}.$$

Transferring this constant to the right-hand side gives the standard form

$$\left(x+\dfrac12\right)^{2} + \left(y+\dfrac34\right)^{2} = \dfrac{5}{16}.$$

Thus the locus is a circle with centre $$\left(-\dfrac12,\;-\dfrac34\right)$$ and radius $$\sqrt{\dfrac{5}{16}}=\dfrac{\sqrt5}{4}.$$ Its diameter is twice the radius, namely

$$2\times\dfrac{\sqrt5}{4}=\dfrac{\sqrt5}{2}.$$

This matches the description of Option D: “circle whose diameter is $$\dfrac{\sqrt{5}}{2}$$.”

Hence, the correct answer is Option D.