a) Ta có : \(-2a = -2 \Rightarrow a = 1\)
\(-2b = -2 \Rightarrow b = 1 \Rightarrow I(1; 1)\)
\({R^2} = {a^2} + {b^2} - c \)\(= {1^2} + {1^2} - ( - 2) = 4 \Rightarrow R = \sqrt 4 = 2\)
b) \(16{x^2} + {\rm{ }}16{y^2} + {\rm{ }}16x{\rm{ }}-{\rm{ }}8y{\rm{ }}-{\rm{ }}11{\rm{ }} = {\rm{ }}0\)
\( \Leftrightarrow {x^2} + {y^2} + x - {1 \over 2}y - {{11} \over {16}} = 0\)
\(\eqalign{
& - 2a = 1 \Rightarrow a = - {1 \over 2} \cr
& - 2b = - {1 \over 2} \Rightarrow b = {1 \over 4} \cr
& \Rightarrow I\left( { - {1 \over 2};{1 \over 4}} \right) \cr} \)
\({R^2} = {a^2} + {b^2} - c \)\(= {\left( { - {1 \over 2}} \right)^2} + {\left( {{1 \over 4}} \right)^2} - \left( { - {{11} \over {16}}} \right) = 1\)\( \Rightarrow R = \sqrt 1 = 1\)
c)
\(\eqalign{
& - 2a = - 4 \Rightarrow a = 2 \cr
& - 2b = 6 \Rightarrow b = - 3 \cr
& \Rightarrow I\left( {2; - 3} \right) \cr} \)
\({R^2} = {a^2} + {b^2} - c \)\(= {2^2} + {\left( { - 3} \right)^2} - \left( { - 3} \right) = 16 \)\(\Rightarrow R = \sqrt {16} = 4\)
