Complete the square of the equations in this part to render them in a form of the equation of a circle.

${\displaystyle x^{2}+2x+y^{2}=5\,}$

${\displaystyle x^{2}+2x=5-y^{2}\,}$

${\displaystyle x^{2}+2x+1=6-y^{2}\,}$

We add one to both sides so we can write the right hand side as a single power.

${\displaystyle (x+1)^{2}=6-y^{2}\,}$

${\displaystyle (x+1)^{2}+y^{2}=6\,}$

${\displaystyle x^{2}+4x+y^{2}-4y=20\,}$

${\displaystyle x^{2}+4x=-(y^{2}-4y)+20\,}$

${\displaystyle x^{2}+4x+4=-(y^{2}-4y)+20\,}$

${\displaystyle (x+2)^{2}=-(y^{2}-4y)+24\,}$

${\displaystyle y^{2}-4y=-(x+2)^{2}+24\,}$

${\displaystyle y^{2}-4y+4=-(x+2)^{2}+28\,}$

${\displaystyle (y-2)^{2}+(x+2)^{2}=28\,}$

Multiply both sides of the inequality by both numerator and denominator of both sides to get:

${\displaystyle (1+t^{2})(1-s^{2})>(1-t^{2})(1+s^{2})\,}$

Then, expand and simplify:

${\displaystyle 1-s^{2}+t^{2}-(ts)^{2}>1+s^{2}-t^{2}-(ts)^{2}\,}$

${\displaystyle -s^{2}+t^{2}>s^{2}-t^{2}\,}$

${\displaystyle -2s^{2}+t^{2}>-t^{2}\,}$

${\displaystyle -2s^{2}>-2t^{2}\,}$

${\displaystyle s^{2}<t^{2}\,}$

${\displaystyle s<t\,}$