The equation of circle of minimum radius...

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 :

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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 :

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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 :

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