(4x^(2)-3y^(2)):(2x^(2)+5y^(2))=12:19," then "x:y

Solve for (x - 1)^(2) and (y + 3)^(2) , 2x^(2) - 5y^(2) - x - 27y - 26 = 3(x + y + 5) and 4x^(2) - 3y^(2) - 2xy + 2x - 32y - 16 = (x - y + 4)^(2) .

If (5x - 3y) : (2x + 4y) = 11 : 12 , then find x : y .

Find the radical center of the circles x^(2)+y^(2)+4x+6y=19,x^(2)+y^(2)=9,x^(2)+y^(2)-2x-4y=5

Find each of the following products: (i) (4x + 5y) (4x - 5y) (ii) (3x^(2) + 2y^(2)) (3x^(2) - 2y^(2))

The radical centre of the circles x^(2)+y^(2)=9 , x^(2)+y^(2)-2x-2y-5=0 , x^(2)+y^(2)+4x+6y-19=0 is A) (0,0) (B) (1,1) (C) (2,2) (D) (3,3)

Simplify : (2x+y)^(5)-5y(2x+y)^(4)+10y^(2)(2x+y)^(3)-10y^(3)(2x+y)^(2)+5y^(4)(2x+y)-y^(5)

If A_(1),A_(2),A_(3) be the areas of circles x^(2)+y^(2)+4x+6y-19=0, x^(2)+y^(2)=9, x^(2)+y^(2)-4x-6y-12=0 respectively then