The two curves `2x^2+y^2=20,x^2-4y^2+8=0`

Find the angle between the two curves y=2x^2 and y=x^2+4x-4

If the two curves 2x^(2)+kxy+y^(2)+2x+4y+1=0 and x^(2)+4xy+4y^(2)-x+2y+4=0 intersect at 4 non concyclic points then k=

The common tangent at the point of contact of the two circles x^2+y^2-2x-4y-20=0, x^2+y^2+6x+2y-90=0 is

The common tangent at the point of contact of the two circles x^(2)+y^(2)-2x-4y-20=0 and x^(2)+y^(2)+6x+2y-90=0 is

The common tangent at the point of contact of the two circles x^2+y^2-2x-4y-20=0, x^2+y^2+6x+2y-90=0 is

The two circles x^2+y^2-2x-3=0 and x^2+y^2-4x-6y-8=0 are such that :

A curve satisfies the differential equation (dy)/(dx)=(x+1-xy^2)/(x^2y-y) and passes through (0,0) (1) The equation of the curve is x^2+y^2+2x=x^2y^2 (2) The equation of the curve is x^2+y^2+2x+2y=x^2y^2 (3) x=0 is a tangent to curve (4) y=0 is a tangent to curve

The length of the common chord of the two circles x^2+y^2-4y=0 and x^2+y^2-8x-4y+11=0 is

The length of the common chord of the two circles x^2+y^2-4y=0 and x^2+y^2-8x-4y+11=0 is