Two circles have the equations` x^2+y^2+lambdax+c=0 and x^2+y^2+mux+c=0`. Prove that one of the circles will be within the other if `lambdamugt0 and cgt0`.

Two circles have equations x^(2)+y^(2)-4x-6y-7=0 and x^(2)+y^(2)-2x-3=0 then

x^(2) + y^(2) + 6x = 0 and x^(2) + y^(2) – 2x = 0 are two circles, then

The circle represented by the equation x ^(2) + y^(2) + 2gx + 2fy + c=0 will be a point circle, if

Two circles x^(2) + y^(2) - 2x - 4y = 0 and x^(2) + y^(2) - 8y - 4 = 0

If one of the circles x^(2)+y^(2)+2ax+c=0 and x^(2)+y^(2)+2bx+c=0 lies within the other,then

If the circles x^2+y^2+2ax+c=0 and x^2+y^2+2by+c=0 touch each other , prove that c is always positive .

If two circles x^(2) + y^(2) - 2ax + c = =0 and x^(2) + y^(2) - 2by + c = 0 touch each other , then c =

The circles whose equations are x^(2)+y^(2)+10x-2y+22=0 and x^(2)+y^(2)+2x-8y+8=0 touch each other. The circle which touch both circles at, the point of contact and passing through (0,0) is, 1) 9(x^(2)+y^(2))-15x-20y=0, 2) 5(x^(2)+y^(2))-18x-80y=0 3) 7(x^(2)+y^(2))-18x-80y=0, 4) x^(2)+y^(2)-9x-40y=0 ]]