If x is the central dot, y1 to y6 the dots participating in 3 lines, z1 to z12 the dots only participating in one line, and if A=y1+y2+y3+y4+y5+y6 and B=z1+z2+...+z12, then:

x+A+B=190 (the sum of all natural numbers from 1 to 19)
6x+3A+B=264 (the sum of all 12 lines with a sum of 22)

Therefore:
A=(74-5x)/2 (1)
B=(306+3x)/2 (2)
With x,A,B being natural numbers

Applying to x its possible values, i.e. 1,2,3,...,19 we have:
For x=1, A= 34,5 --> so x isn't 1
All odd values for x give the same result

For x=2, A=32 and B=168. We accept x=2, because in this case A has a minimum value of A=1+3+4+5+6+7=26

Same goes for x=4, since A=27 and B=159, with Amin=24

For x=6, A=22 and B=162, with Amin=22; x=6 is the last acceptable value for x, since:
for x=8, even though A and B are natural numbers, and therefore acceptable values, A < Amin and therefore x>8 is not an acceptable solution. Moreover, x has been shown to be an even number.

However, if x=6, one cannot find any solution when trying to solve the grid. Therefore, solutions are acceptable only if x=2 or x=4.