Lời giải chi tiết:
Ta có: (overrightarrow {{n_d}} = left( {2;,,1} right).)
Đường tròn (left( C right)) có tâm (Ileft( {1; - 1} right)) và bán kính (R = sqrt {10} .)
Giả sử tiếp tuyến (Delta :,,,ax + by + c = 0,,,left( {{a^2} + {b^2} ne 0} right) Rightarrow overrightarrow {{n_Delta }} = left( {a;,,b} right).)
(Delta ) là tiếp tuyến của (left( C right) Rightarrow dleft( {I;,,Delta } right) = R = sqrt {10} .)
( Leftrightarrow frac{{left| {a - b + c} right|}}{{sqrt {{a^2} + {b^2}} }} = sqrt {10} Leftrightarrow left| {a - b + c} right| = sqrt {10left( {{a^2} + {b^2}} right)} ,,,,left( 1 right))
Theo đề bài ta có: (Delta ) tạo với (d) một góc ({45^0} Rightarrow cos {45^0} = frac{{left| {overrightarrow {{n_Delta }} .overrightarrow {{n_d}} } right|}}{{left| {overrightarrow {{n_Delta }} } right|.left| {overrightarrow {{n_d}} } right|}})
(begin{array}{l} Leftrightarrow frac{{left| {2a + b} right|}}{{sqrt {{2^2} + 1} .sqrt {{a^2} + {b^2}} }} = frac{{sqrt 2 }}{2} Leftrightarrow 2left| {2a + b} right| = sqrt {10} .sqrt {{a^2} + {b^2}} Leftrightarrow 4{left( {2a + b} right)^2} = 10left( {{a^2} + {b^2}} right) Leftrightarrow 6{a^2} + 16ab - 6{b^2} = 0 Leftrightarrow 3{a^2} + 8ab - 3{b^2} = 0 Leftrightarrow left( {a + 3b} right)left( {3a - b} right) = 0 Leftrightarrow left[ begin{array}{l}a + 3b = 03a - b = 0end{array} right. Leftrightarrow left[ begin{array}{l}a = - 3ba = frac{b}{3}end{array} right.end{array})
+) Với (a = - 3b Rightarrow left( 1 right) Leftrightarrow left| { - 3b - b + c} right| = sqrt {10left( {9{b^2} + {b^2}} right)} )
(begin{array}{l} Leftrightarrow left| {c - 4b} right| = sqrt {100{b^2}} Leftrightarrow left| {c - 4b} right| = 10left| b right| Leftrightarrow left[ begin{array}{l}c - 4b = 10bc - 4b = - 10bend{array} right. Leftrightarrow left[ begin{array}{l}c = 14bc = - 6bend{array} right. Rightarrow left[ begin{array}{l}left{ begin{array}{l}a = - 3bc = 14bend{array} right. Rightarrow {Delta _1}:,,, - 3x + y + 14 = 0left{ begin{array}{l}a = - 3bc = - 6bend{array} right. Rightarrow {Delta _2}:,, - 3x + y - 6 = 0end{array} right. Rightarrow left[ begin{array}{l}{Delta _1}:,,,3x - y - 14 = 0{Delta _2}:,,3x - y + 6 = 0end{array} right..end{array})
+) Với (a = frac{b}{3} Rightarrow left( 1 right) Leftrightarrow left| {frac{b}{3} - b + c} right| = sqrt {10left( {frac{{{b^2}}}{9} + {b^2}} right)} )
(begin{array}{l} Leftrightarrow left| {c - frac{{2b}}{3}} right| = sqrt {frac{{100{b^2}}}{9}} Leftrightarrow left| {c - frac{{2b}}{3}} right| = frac{{10}}{3}left| b right| Leftrightarrow left[ begin{array}{l}c - frac{{2b}}{3} = frac{{10b}}{3}c - frac{{2b}}{3} = - frac{{10b}}{3}end{array} right. Leftrightarrow left[ begin{array}{l}c = 4bc = - frac{8}{3}bend{array} right.end{array})
( Rightarrow left[ begin{array}{l}left{ begin{array}{l}a = frac{b}{3}c = 4bend{array} right. Rightarrow {Delta _3}:,,,frac{1}{3}x + y + 4 = 0left{ begin{array}{l}a = frac{b}{3}c = - frac{{8b}}{3}end{array} right. Rightarrow {Delta _4}:,,frac{1}{3}x + y - frac{8}{3} = 0end{array} right.)
( Rightarrow left[ begin{array}{l}{Delta _3}:,,,x + 3y + 12 = 0{Delta _4}:,,x + 3y - 8 = 0end{array} right..)
Như vậy có 4 tiếp tuyến thỏa mãn bài toán.
Chọn A.
