Let the linear l equation be y=kx+b,

The straight line l intersects with the point P (2 2,3)

∴3=2k+b, that is, b=3-2k,

The straight line l is kx-y+3-2k = 0...( 1)

The straight line l is tangent to c: x2+y2=4,

Point c (0 0,0), r=2

∴d=|3-2k|/(√k？ + 1)=2

Deformation: (3-2k)? =4(k？ + 1)

∴k=5/ 12

Substitute (1) to get:

The equation of the straight line L is 5x- 12y+26=0.