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.