Let us call the circle #O#, whose equation is #x^2+y^2=1#. It has center #(0,0)# and radius #1#. And obviously the point #(0,3)# is outside the circle and hence we will have two tangents.
To construct tangents, we draw a cirle on the line joining its center #(0,0)# and #(0,3)# forming the diameter. Let us call this circle as #P#. As the size of diameter is #3#, its radius is #3/2# and center would be midpoint between #(0,0)# and #(0,3)# i.e. #(0,3/2)#. Hence, its equation is #x^2+(y-3/2)^2=(3/2)^2# or
#x^2+y^2-3y=0#
And the tangents will touch the circle at the points of intersection of the two circles and these points can be obtained by solving the two simultaneous equations #x^2+y^2=1# and #x^2+y^2-3y=0#.
Putting value of #x^2+y^2=1# in to second equation, we get
#1-3y=0# or #y=1/3#
Putting this in #x^2+y^2=1#, we get #x=+-sqrt(1-1/9)=+-(2sqrt2)/3#
Hence two points are #((2sqrt2)/3,1/3)# and #(-(2sqrt2)/3,1/3)#
and joining them with #(0,3)# gives the equation of two tangents
i.e. #(y-3)/(1/3-3)=(x-0)/((2sqrt2)/3-0)#
or #2sqrt2(3y-9)=3x(1-9)# or #24x+6sqrt2y-18sqrt2=0#
or #4x+sqrt2y-3sqrt2=0#
and #(y-3)/(1/3-3)=(x-0)/(-(2sqrt2)/3-0)#
or #-2sqrt2(3y-9)=3x(1-9)# or #24x-6sqrt2y-18sqrt2=0#
or #4x-sqrt2y+3sqrt2=0#
graph{(x^2+y^2-3y)(x^2+y^2-1)(4x+sqrt2y-3sqrt2)(4x-sqrt2y+3sqrt2)=0 [-4.96, 5.04, -0.98, 4.02]}
