[" 101.The minimum radius of the circle which contains the three "],[" circles,"x^(2)+y^(2)-4y-5=0,x^(2)+y^(2)+12x+4y+31=0" and "],[x^(2)+y^(2)+6x+12y+36=0" ,is "]

Find the equation of the circle of minimum radius which contains the three cricles x^(2)-y^(2)-4y-5=0 x^(2)+y^(2)+12x+4y+31=0 and x^(2)+y^(2)+6x+12y+36=0

Find the equation of the circle of minimum radius which contains the three circles x^2-y^2-4y-5=0 x^2+y^2+12x+4y+31=0 and x^2+y^2+6x+12y+36=0

The equation of the circle which cuts the three circles x^(2)+y^(2)-4x-6y+4=0 , x^(2)+y^(2)-2x-8y+4=0 and x^(2)+y^(2)-6x-6y+4=0 orthogonally is

The equation of circle of minimum radius which contacts the three circle x^(2)+y^(2)-4y-5=0,x^(2)+y^(2)+12x+4y+31=0,x^(2)+y^(2)+6x+12y+36=0 then the radius of given circle is (l+(m)/(36)sqrt(949)) then the value of 1+m is :

The equation of a circle which cuts the three circles x^(2)+y^(2)+2x+4y+1=0,x^(2)+y^(2)-x-4y+8=0 and x^(2)+y^(2)+2x-6y+9=0 orthogonlly is

The circles x^(2)+y^(2)+6x+6y=0 and x^(2)+y^(2)-12x-12y=0

The equation of circle of minimum radius which contacts the three circle x^2 + y^2 -4y-5 = 0, x^2 +y^2 +12x +4y +31 = 0, x^2 +y^2 + 6x +12y + 36 = 0 then the radius of given circle is (l + m/36sqrt949) then the value of l + m is :

Prove that the centres of the three circles x^(2)+y^(2)-4x6y12=0,x^(2)+y^(2)+2x+4y-5=0 and x^(2)+y^(2)-10x16y+7=0 are collinear.

The radical centre of the circles x^(2) + y^(2)- 2x + 6y = 0 , x^(2) + y^(2) - 4x - 2y + 6 = 0 , x^(2) + y^(2) - 12x + 12y + 30 = 0 is

The centre of circle cutting the three circles x^(2)+y^(2)-4x-4y+5=0 , x^(2)+y^(2)-4x-6y+5=0 , x^(2)+y^(2)-4x-8y+5=0 orthogonally is