Circles

Let BC  = x

Triangles AEC  and BDC  are similar

BC / BD  = AC / AE

x / 1 = (4 + x) / 3

3x = 4 + x

2x = 4

x = 2 =  BC

{ DC = sqrt (BC^2 - BD^2)  = sqrt ( 2^2 - 1^2)  = sqrt (3)

tan(angle DCB)  =  - ( BD / DC)  = -1/sqrt (3)

Equation of tangent line       : y = (-1/sqrt 3) ( x -6)   }