`x^2-y^2+2x+1=?`

The angle between the lines x^(2)-y^(2)-2 x-1=0 is

Consider the two curves C_1 ; y^2 = 4x , C_2 : x^2 + y^2 - 6x + 1 = 0 then :

Consider the two curves C_1:y^2=4x,C^2:x^2-y^2-6x+1=0. Then.

A circle of constant radius a passes through the origin O and cuts the axes of coordinates at points P and Q . Then the equation of the locus of the foot of perpendicular from O to P Q is (x^2+y^2)(1/(x^2)+1/(y^2))=4a^2 (x^2+y^2)^2(1/(x^2)+1/(y^2))=a^2 (x^2+y^2)^2(1/(x^2)+1/(y^2))=4a^2 (x^2+y^2)(1/(x^2)+1/(y^2))=a^2

For the circles 3x^2+3y^2+x+2y-1=0 , 2x^2+2y^2+2x-y-1=0 radical axis is