" (b) "x^(2)+y^(2)=sqrt(2)a^(2)" And "x^(2)-y^(2)=a^(2)

if a>2b>0, then positive value of m for which y=mx-b sqrt(1+m^(2)) is a common tangent to x^(2)+y^(2)=b^(2) and (x-a)^(2)+y^(2)=b^(2) is

The equation of a circle of radius 1touching the circles x^(2)+y^(2)-2|x|=0 is x^(2)+y^(2)+2sqrt(2)x+1=0x^(2)+y^(2)-2sqrt(3)y+2=0x^(2)+y^(2)+2sqrt(3)y+2=0x^(2)+y^(2)-2sqrt(2)+1=0

Simplify : (sqrt(x^(2)+y^(2))-y)/(x-sqrt(x^(2)-y^(2))) div (sqrt(x^(2)-y^(2))+x)/(sqrt(x^(2)+y^(2))+y)

x(dy)/(dx)-y=2sqrt(y^(2)-x^(2))

Simplify : (sqrt(x^(2)+y^(2))-y)/(x-sqrt(x^(2)-y^(2)))-:(sqrt(x^(2)-y^(2))+x)/(sqrt(x^(2)+y^(2))+y)

Simplify : (sqrt(x^(2)+y^(2))-y)/(x-sqrt(x^(2)+y^(2)))-:(sqrt(x^(2)+y^(2))+x)/(sqrt(x^(2)+y^(2))+y)

If y - mx = sqrt(a^(2) m^(2) + b^(2)) and x + my = sqrt(a^(2) + b^(2) m^(2)) , then prove that x^(2) + y^(2) = a^(2) + b^(2)

Rationalise the denominator: (a) (1)/(root(3)(3) + root(3)(2)) , (b) (2)/(sqrt5 + sqrt3 + sqrt2) , (c) (x^(2))/(sqrt(x^(2) + y^(2)) - y) , (d) (1)/(sqrt6 + sqrt5 - sqrt11) (e) (sqrt(x + 2y) - sqrt(x -2y))/(sqrt(x + 2y) + sqrt(x - 2y)) , (f) (sqrt10 + sqrt5 - sqrt3)/(sqrt10 - sqrt5 + sqrt3)

The equation sqrt([(x-2)^(2)+y^(2)])+ sqrt([(x+2)^(2)+y^(2)])=4

The eccentricity of the conic represented by sqrt((x+2)^(2)+y^(2))+sqrt((x-2)^(2)+y^(2))=8 is