Transforming the given equation of the circle #3x²+3y²+12x+8y+8=0# in standard form we get.
#3x²+3y²+12x+8y+8=0#
#=>x²+y²+4x+8/3y+8/3=0#
#=>(x²+2*x*2+ 2^2)+y^2+2*y*4/3+(4/3)^2=4+16/9-8/3#
#=>(x+2)^2+(y+4/3)^2=(sqrt(28/9))^2#
So the center of the circle C #to(-2,-4/3)#
and radius of the circle #rtosqrt(28/9)#
If #PandQ# are points of contact of two tangents drawn from origin #O# to the circle then #DeltasPOCandQOC# will be two congruent right triangles both having common hypotenuse OC.
So #PC=QC=r#
Now #OC^2=(0-(-2))^2+(0-(-4/3))^2#
#=>OC=sqrt(52/9)#
So length of each tangent
#OP=sqrt(OC^2-CP^2)#
#=sqrt(52/9-28/9)#
#=sqrt(8/3)=2sqrt(2/3)=(2sqrt6)/3#
Now #sinanglePOC=(PC)/(OC)#
#=>sinanglePOC=sqrt(28/9)/sqrt(52/9)#
#=>anglePOC=sin^-1(sqrt(7/13))=47.2^@#
So angle between the tangents will be
#anglePOQ=2*anglePOC=94.4^@#
But #angle PCQ=2*angle PCO=2*(90^@-47.2^@)=85.6^@#