Let the equation of line is 

      y = mx + c ...............1

=> y - mx - c = 0 ...........2

Given distance from origin to this line  = 2

=> |(0 - 0 - c)/√(1+m² )| = 2

=> c/√(1+m² ) = 2

Square on both side, we get

      c² /(1+m² ) = 4

=> c² = 4(1+m² ) ......3

Again given line passes through (2,2)

So   2 - 2m - c = 0

= > c = 2-2m ..........4

Put this value in equation3, we get

     (2 - 2m)² = 4(1+m² )

=> 4 + 4m² - 8m = 4 + 4m²

=> 8m = 0

=> m = 0

Put this value in equation 4 , we get

      c= 2 - 0

=> c = 2

Now Put this value of c and m in equation 1 , we get

y = 0*x + 2

=> y = 2

This is the required equation of line.