(3x+4y+1)^2+(x+y+3)^2=0

On expanding we get,

4x.x + 5y.y + 13xy + 7y + 6x + 10=0

it is in the form - ax.x + by.y + 2hxy + 2gx + 2fy + c=0.

to check we first find the area i.e.   abc + 2fgh - af.f - bg.g - ch.h= area.

it is non zero,  hence it does not represent the straight lines.

now, we check, h.h= 169/4 and a.b= 20.

here, h.h>a.b ;  hence, the equation represents the hyperbola with center  (-11, 8).