Finding the axes of an ellipse from deformation of a circle

Question

Consider a circle inscribed in a square (radius of circle equals half the height of the square). This figures are deformed so that the square is now a parallelogram with height equal to the original, and its base parallel and equal in length to the base of the original square. The inclined side of the parallelogram forms an angle whose tangent is $\gamma$ with the line normal to the base (a vertical line). The circle deforms into an ellipse, tangent to the parallelogram at its side's midpoints, with semiaxes $a>b$. I know that the relationship between the semiaxes lenghts and the angle $\gamma$ can be obtained from the following equation: $$ a^2,b^2 = \frac{\gamma^2+2±\gamma\sqrt{\gamma^2+4}}{2}\space\space\space\space equation \space1 $$ In an old text I found the forms: $$ \gamma=\frac{a-b}{\sqrt{ab}}\space\space\space\space equation\space 2 $$ and $$ a,b=\frac{±\gamma+\sqrt{\gamma^2+4}}{2}\space\space\space\space equation\space 3 $$ But I can't get algebraically from equation 1 to any of the other two, even if when I try them with the same parameters gives the same results (example, setting a value for $\gamma$ and obtaining $a$ and $b$ in equation 1, gives the same $a$ and $b$ in equation 3; and using these values for $a$ and $b$ in equation 2 gives the correct value for $\gamma$). Does anyone can relate these equations algebraically so that I am sure that I can use them indistinctly? Thanks