A B → = ( − 7 ; − 1 ) \overrightarrow{AB}=(-7;-1) A B = ( − 7 ; − 1 ) n → = ( 1 ; − 7 ) ⇒ \overrightarrow{n}=(1;-7)\Rightarrow n = ( 1 ; − 7 ) ⇒ 1 ( x − 4 ) − 7 ( y − 2 ) = 0 ⇔ x − 7 y + 10 = 0 1(x - 4) - 7(y - 2) = 0\Leftrightarrow x-7y+10=0 1 ( x − 4 ) − 7 ( y − 2 ) = 0 ⇔ x − 7 y + 1 0 = 0
C ∈ ( d ) ⇒ C ( c ; − c ) ( c > 0 ) C \in (d)\Rightarrow C(c;-c)(c> 0) C ∈ ( d ) ⇒ C ( c ; − c ) ( c > 0 )
⇒ d ( C , A B ) = ∣ c + 7 c + 10 ∣ 1 2 + 7 2 = ∣ 8 c + 10 ∣ 50 ; A B = 50 \Rightarrow d(C,AB)=\frac{\left | c+7c+10 \right |}{\sqrt{1^{2}+7^{2}}}=\frac{\left | 8c+10 \right |}{\sqrt{50}};AB=\sqrt{50} ⇒ d ( C , A B ) = 1 2 + 7 2 ∣ c + 7 c + 1 0 ∣ = 5 0 ∣ 8 c + 1 0 ∣ ; A B = 5 0
Diện tích tam giác ABC bằng 25 nên ta có
S △ A B C = 1 2 d ( C , A B ) = ∣ 8 c + 10 ∣ 2 50 . 50 = 25 ⇔ [ c = 5 c = − 15 2 < 0 ⇒ C ( 5 ; − 5 ) S_{\triangle ABC}=\frac{1}{2}d(C,AB)=\frac{\left | 8c+10 \right |}{2\sqrt{50}}.\sqrt{50}=25\Leftrightarrow \bigg \lbrack\begin{matrix} c=5\\c=-\frac{15}{2}< 0 \end{matrix}\Rightarrow C(5;-5) S △ A B C = 2 1 d ( C , A B ) = 2 5 0 ∣ 8 c + 1 0 ∣ . 5 0 = 2 5 ⇔ [ c = 5 c = − 2 1 5 < 0 ⇒ C ( 5 ; − 5 )
Gọi (C) là đường tròn ngoại tiếp tam giác ABC có phương trình là:
( C ) : x 2 + y 2 − 2 a x − 2 b y + c = 0   ( a 2 + b 2 − c > 0 ) (C):x^{2}+y^{2}-2ax-2by+c=0\; (a^{2}+b^{2}-c> 0) ( C ) : x 2 + y 2 − 2 a x − 2 b y + c = 0 ( a 2 + b 2 − c > 0 )
Do A, B, C nằm trên (C) nên ta có hệ
{ 4 2 + 2 2 − 8 a − 4 b + c = 0 ( − 3 ) 2 + 1 2 + 6 a − 2 b + c = 0 5 2 + ( − 5 ) 2 − 10 a + 10 b + c = 0 ⇔ { − 8 a − 4 b + c = − 20 6 a − 2 b + c = − 10 − 10 a + 10 b + c = − 50 \left\{\begin{matrix} 4^{2}+2^{2}-8a-4b+c=0\\(-3)^{2}+1^{2}+6a-2b+c=0 \\5^{2}+(-5)^{2}-10a+10b+c=0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -8a-4b+c=-20\\6a-2b+c=-10 \\-10a+10b+c=-50 \end{matrix}\right. ⎩ ⎨ ⎧ 4 2 + 2 2 − 8 a − 4 b + c = 0 ( − 3 ) 2 + 1 2 + 6 a − 2 b + c = 0 5 2 + ( − 5 ) 2 − 1 0 a + 1 0 b + c = 0 ⇔ ⎩ ⎨ ⎧ − 8 a − 4 b + c = − 2 0 6 a − 2 b + c = − 1 0 − 1 0 a + 1 0 b + c = − 5 0
⇔ { a = 1 b = − 2 c = − 20 ⇒ \Leftrightarrow \left\{\begin{matrix} a=1\\b=-2 \\c=-20 \end{matrix}\right.\Rightarrow ⇔ ⎩ ⎨ ⎧ a = 1 b = − 2 c = − 2 0 ⇒ x 2 + y 2 − 2 x + 4 y − 20 = 0 x^{2}+y^{2}-2x+4y-20=0 x 2 + y 2 − 2 x + 4 y − 2 0 = 0
