If the two circles $${\left( {x - 1} \right)^2} + {\left( {y - 3} \right)^2} = {r^2}$$     and $${x^2} + {y^2} - 8x + 2y + 8 = 0$$      intersect in two distinct point, then-

Solution :
$$\eqalign{ & \left| {{r_1} - {r_2}} \right| < {C_1}{C_2}{\text{ for intersection}} \cr & \Rightarrow r - 3 < 5 \Rightarrow r < 8.....\left( 1 \right) \cr & {\text{and }}{r_1} + {r_2} > {C_1}{C_2},\,\,\,r + 3 > 5 \Rightarrow r > 2.....\left( 2 \right) \cr & {\text{From }}\left( 1 \right){\text{ and }}\left( 2 \right),{\text{ we get}} \cr & 2 < r < 8 \cr} $$