(" iij ")-4y^(2)+3y+5=0

Solution of D.E (dy)/(dx)=(2x+5y)/(2y-5x+3) is,if (y(0)=0) (1) x^(2)-y^(2)+5xy-3y=0 (2) x^(2)+y^(2)+5xy-3y=0 (3) x^(2)-y^(2)+5xy+3y=0 (4) x^(2)-y^(2)-5xy-3y=0

Subtract : 5y^(4) - 3y^(3) + 2y^(2) + y - 1 from 4y^(4) - 2y^(3) - 6y^(2) - y + 5

The radical axis of circles x^(2) + y^(2) + 5x + 4y - 5 = 0 and x^(2) + y^(2) - 3x + 5y - 6 = 0 is

the radical axis of the circles x^(2) + y^(2) + 3x + 4y - 5 = 0 " and " x^(2) + y^(2) - 5x + 5y - 6 = 0 is

the slope of the radical axis of the circles x^(2) + y^(2) + 3x + 4y - 5 = 0 " and " x^(2) + y^(2) - 5x + 5y - 6 = 0 is

If the point of intersection of kx+4y+2=0, x-3y+5=0 lies on 2x+7y-3=0 then k =

Find the value of k if kx+ 3y - 1 = 0, 2x + y + 5 = 0 are conjugate lines with respect to the circle x^(2) + y^(2) - 2x - 4y - 4 =0 .

Find the value of k if kx+ 3y - 1 = 0, 2x + y + 5 = 0 are conjugate lines with respect to the circle x^(2) + y^(2) - 2x - 4y - 4 =0 .