`x^2+y^2=4`
`2x^2+3y^2=12`

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

If (2x^2-y^2)^4=256 and (x^2+y^2)^5=243 , then find x^4-y^4 The following steps are involved in solving the above problem. Arragne then in sequential order. (A) (x^2-y^2)^4=256=4^4 and (x^2+y^2)^5=3^5 (B) x^4-y^4=12 (C) (x^2-y^2)(x^2+y^2)=4xx3 (D) x^2-y^2=4 and x^2+y^2=3

Prove that the radi of the circles x^(2)+y^(2)=1x^(2)+y^(2)-2x-6y=6 and x^(2)+y^(2)-4x-12y=9 are in arithmetic progression.

If a parabola is represented by 25(x^(2)+y^(2))=(4x-3y-12)^(2), then

(3) x+2y=-1;2x-3y=12

From the sum of x^2+3y^2-6x y ,2x^2-y^2+8x y , y^2+8 and x^2-3x y\ subtract -3x^2+4y^2-x y+x-y+3.