" afic "4x^(2)-3y^(2):2x^(2)+5y^(2)=12:19" at at "x:y=?

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

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)

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

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

Identify the type of conic section for each of the equations 1. 2x^(2) -y^(2) = 7 2. 3x^(2) +3 y^(2) -4x + 3y + 10 =0 3. 3x^(2) + 2y^(2) = 14 4. x^(2) + y^(2) + x-y=0 5. 11x^(2) -25y^(2) -44x + 50y - 256 =0 6. y^(2) + 4x + 3y + 4=0

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

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) .