How can I get the center and radius of this circle?

Here's a one-liner for obtaining an implicit Cartesian equation:

circ = First[GroebnerBasis[ComplexExpand[Abs[(x + I y + 1)/(x + I y - 1)] == a,
                                         TargetFunctions -> {Re, Im}], {x, y, a}]]
   -1 + a^2 - 2 x - 2 a^2 x - x^2 + a^2 x^2 - y^2 + a^2 y^2

From here, we can use the technique from this answer:

vars = {x, y};
{cnst, lin, quad} = MapAt[Diagonal, Normal[CoefficientArrays[circ, vars]], {3}];
cnst + Total[MapThread[depress[#1 FromDigits[{##2}, #1]] &, {vars, quad, lin}]]
   -1 + a^2 - (1 + a^2)^2/(-1 + a^2) +
   (-1 + a^2) (-((2 + 2 a^2)/(2 (-1 + a^2))) + x)^2 - y^2 + a^2 y^2

Manual massaging of this result leads to the form

(x + (1 + a^2)/(1 - a^2))^2 + y^2 == (4 a^2)/(1 - a^2)^2

which means the result is a circle with center {(a^2 + 1)/(a^2 - 1), 0} and radius Abs[2 a/(1 - a^2)].

For concrete values of a this can be done as follows.

a=9/10;d = ImplicitRegion[ComplexExpand[Abs[(x + I y + 1)/(x + I y - 1)]] == a, {x, y}];
c = RegionCentroid[d]

{-(181/19), 0}

r=RegionDistance[d, c]

180/19

Something really cool I learned, but I cannot find the reference. Probably from one of the authors of Indra's Pearls...

There is a one-to-one correspondence between circles and Hermitian matrices of negative determinant. Thus, any circle may be represented by the Hermitian matrix

$H = \left[ \begin{array}{cc} 1 & -p \\ -p^* & |p|^2-r^2 \end{array} \right]$

where the complex number $p$ is the circle centre, and the real number $r$ is the circle radius.

The mapping of an input circle to an output circle is accomplished by $G=(M^{-1})^{T*} \cdot H \cdot M^{-1}$, where $M=\{\{a,b\},\{c,d\}\}$ in the Mobius transform. In component form,

$ G = \left[ \begin{array}{cc} d^* & -c^* \\ -b^* & a^* \end{array} \right] \cdot \left[ \begin{array}{cc} 1 & -p \\ -p^* & |p|^2-r^2 \end{array} \right] \cdot \left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right] $

where the superscript $*$ indicates complex conjugation. The result is another Hermitian matrix

$ G = \left[ \begin{array}{cc} A & B \\ B^* & C \end{array} \right] = \left[ \begin{array}{cc} 1 & -q \\ -q^* & |q|^2-s^2 \end{array} \right] $

corresponding to a new circle with centre $q$ and radius $s$.

MobiusMap finds the coefficients $\{A,B,C\}$ of the Hermitian matrix $G$, and forms the output circle from them. It has a special case when $A=0$, resulting in a line $U x+V y+W=0$. Ratios taken to form the output circle are independent of whether or not the mapping has unit determinant.

MobiusMap[m_?MatrixQ, Circle[{x_, y_}, r_]] :=
   Block[{v = -x - I y, w = x^2 + y^2 - r^2, a, b, c},
      a = Abs[m[[2, 2]]]^2 + Abs[m[[2, 1]]]^2 w - 
          2 Re[m[[2, 1]] Conjugate[m[[2, 2]]] v];
      b = m[[1, 1]] (Conjugate[m[[2, 2]]] v - Conjugate[m[[2, 1]]] w) + 
          m[[1, 2]] (Conjugate[v*m[[2, 1]]] - Conjugate[m[[2, 2]]]);
      c = Abs[m[[1, 2]]]^2 + Abs[m[[1, 1]]]^2 w - 
          2 Re[m[[1, 1]] Conjugate[m[[1, 2]]] v];
      If[
         Chop[N[a]] == 0.,
         LineUVW[{Re[b], Im[b]}, c/2],
         Circle[-{Re[b], Im[b]}/a, Sqrt[b*Conjugate[b] - a*c]/Abs[a]]]
]

Your mapping is $M=\{\{1,1\},\{1,-1\}\}$. The Hermitian matrix corresponding to the mapped circle is $G=\{\{1 - r^2, 1 + r^2\}, \{1 + r^2, 1 - r^2\}\}$. Thus, the centre $q=-\{Re[B],Im[B]\}/A=\frac{r^2+1}{r^2-1}$, and the radius $s=\frac{2r}{Abs[1-r^2]}$.

Manipulate[
   Module[{circle, q, s},
      circle = MobiusMap[{{1, 1}, {1, -1}}, Circle[{0, 0}, r]];
      q = circle[[1]];
      s = circle[[2]];
      Graphics[{Thick, PointSize[0.015],
         Circle[{0, 0}, r], Point[{0, 0}],
         Red, circle, Point[q]
         }, GridLines -> Automatic, Frame -> True,
         PlotLabel -> "Centre q: "<>ToString[q]<>"    Radius s: "<>ToString[s]]],
   {{r, 0.9, "Radius r"}, 0., 2., Appearance -> "Labeled"}]