Inversion swaps the points X, and X' with respect to the circle K because X' is always on the ray from O through X, and X and X' are related by the equation OX*OX'=R*R, by definition of inverse points. To do this in geogebra, X' is just X dialated by a factor of (R*R)/(OX*OX).
The circle C is a circle through inverse points X, and X', and through an arbitrary point B.
The amazing thing about the circle C is that no matter where X is moved, or how big C or K is, C will cut through K orthogonally!
Let T be the point on C whose tangent line goes through O.
Claim that T is on K.
Proof: The circle C is a circle through three points, X, X', and B, where X and X' are inverse points with respect to circle K, and B is just an arbitrary point. The power of O with respect to circle C is equal to OX * OX' = OT * OT, but since X and X' are inverse points with respect to the circle K, OX * OX' = R*R implies OT * OT = R * R implies OT=R implies T is on K, and C cuts through K at a right angle.

Try moving any of the moveable blue points around.