Odpowiedź :
1. Środek odcinka.
[tex]x_S=\frac{x_A+x_B}{2}=\frac{10+(-10)}{2}=\frac{10-10}{2}=\frac{0}{2}=0\\y_S=\frac{y_A+y_B}{2}=\frac{-8+(-4)}{2}=\frac{-8-4}{2}=\frac{-12}{2}=-6\\S=(0,-6)[/tex]
2. Promień.
[tex]r=|AS|=\sqrt{(x_S-x_A)^2+(y_S-y_A)^2}\\r=\sqrt{(0-10)^2+(-6-(-8))^2}\\r=\sqrt{(-10)^2+(-6+8)^2}\\r=\sqrt{(-10)^2+2^2}\\r=\sqrt{100+4}\\r=\sqrt{104}\\r=\sqrt{4\cdot26}\\r=2\sqrt{26}[/tex]
3. Równanie okręgu.
[tex](x-x_S)^2+(y-y_S)^2=r^2\\(x-0)^2+(y-(-6))^2=(\sqrt{104})^2\\x^2+(y+6)^2=104[/tex]
Odpowiedź:
[tex]A=(10,-8)\ \ \ \ B=(-10,-4)\\\\Obliczamy\ \ dlugo\'s\'c\ \ odcinka\ \ AB\\\\|AB|=\sqrt{(x_{B}-x_{A})^2+(y_{B}-y_{A})^2}=\sqrt{(-10-10)^2+(-4-(-8))^2}=\\\\=\sqrt{(-20)^2+(-4+8)^2}=\sqrt{400+4^2}=\sqrt{400+16}=\sqrt{416}=\sqrt{16\cdot26}=4\sqrt{26}\\\\\\r=\frac{1}{2}\cdot|AB|=\frac{1}{2}\cdot4\sqrt{26}=2\sqrt{26}[/tex]
[tex]Obliczamy\ \ wsp\'olrzedne\ \ \'srodka\ \ odcinka\ \ AB\\\\S_{AB}=\left(\dfrac{x_{A}+x_{B}}{2}\ \ ,\ \ \dfrac{y_{A}+y_{B}}{2}\right)\\\\\\S_{AB}=\left(\dfrac{10+(-10)}{2}\ \ ,\ \ \dfrac{-8+(-4)}{y}\right)\\\\\\S_{AB}=\left(\dfrac{10-10}{2}\ \ \ \ ,\ \ \dfrac{-8-4}{2}\right)\\\\\\S_{AB}=\left(\dfrac{0}{2}\ \ \ \ \ \ ,\dfrac{-12}{2}\right)\\\\\\S_{AB}=(0,-6)[/tex]
[tex]Wz\'or\ \ na\ \ r\'ownanie\ \ okregu\\\\(x-a)^2+(y-b)^2=r^2\\\\(x-0)^2+(y-(-6))^2=(2\sqrt{26})^2\\\\x^2+(y+6)^2=2^2\cdot(\sqrt{26})^2\\\\x^2+(y+6)^2=4\cdot26\\\\\underline{x^2+(y+6)^2=104}[/tex]