If (2, 4) is a point interior to the circle $${x^2} + {y^2} - 6x - 10y + \lambda = 0$$      and the circle does not cut the axes at any point then $$\lambda $$ belongs to the interval :

Solution :
$$\eqalign{ & {2^2} + {4^2} - 6 \times 2 - 10 \times 4 + \lambda < 0 \cr & \Rightarrow \,\lambda - 32 < 0{\text{ or }}\lambda < 32 \cr & {\text{Solving }}y = 0,\,{x^2} + {y^2} - 6x - 10y + \lambda = 0, \cr & {\text{we get }}{x^2} - 6x + \lambda = 0, \cr & {\text{which must have imaginary roots i}}{\text{.e}}{\text{.,}} \cr & 36 - 4\lambda < 0{\text{ i}}{\text{.e}}{\text{., }}\lambda > 9 \cr & {\text{Solving }}x = 0,\,{x^2} + {y^2} - 6x - 10y + \lambda = 0, \cr & {\text{we get }}{y^2} - 10y + \lambda = 0, \cr & {\text{which must have imaginary roots i}}{\text{.e}}{\text{.,}} \cr & 100 - 4\lambda < 0{\text{ i}}{\text{.e}}{\text{., }}\lambda > 25.{\text{ So 25}} < \lambda < 32. \cr} $$