Домашнее задание (Уравнение окружности)

$$\left(х+3\right)^2+\left(у-2\right)^2=25$$  

$$\left(х-3\right)^2+\left(у+2\right)^2=25$$  

$$\left(х+2\right)^2+\left(у-3\right)^2=25$$  

$$\left(х-2\right)^2+\left(у+3\right)^2=25$$  

$$\sqrt[]{х^2+\ \ \ у^2\ \ \ }$$  

$$\sqrt[]{\left(х_2-\ \ х_1\right)^2+\left(у\ _2-\ \ у_1\right)^2}$$  

$$\sqrt[]{\left(х_2\ +\ \ \ х_1\right)^2\ -\left(у\ _2\ +\ у_1\right)^2}$$  

$$\left(х_2\ -\ х_1\right)^2\ +\ \left(у_2\ -\ у_1\right)^2$$  

(-2; 3)

(2; -3)

(-2; -3)

(2; 3)

$$\left(х-2\right)^2+\left(у+1\right)^2=1$$  

$$\left(х-2\right)^2+\left(у-1\right)^2=1$$  

$$\left(х+2\right)^2+\left(у+1\right)^2=1$$  

$$\left(х+2\right)^2+\left(у+1\right)^2=4$$  

$$АВ=\sqrt{\left(-1-3\right)^2+\left(2-5\right)^2}$$

$$АВ=\sqrt{\left(-3-1\right)^2+\left(5-2\right)^2}$$

$$АВ=\sqrt{\left(1-3\right)^2+\left(2+5\right)^2}$$

$$АВ=\sqrt{\left(-1-3\right)^2-\left(2-5\right)^2}$$

С( 2; - 1) ; R = 6

$$С(-2;\ 1);\ R=\sqrt{6}$$  

С( -2; - 1) ; R = 6

$$С(2;-1);\ R=\sqrt{6}$$  

$$(х+5)^2+у^2=\sqrt{3}$$  

$$(х+5)^2-у^2=\sqrt{3}$$  

$$(х-5)^2+у^2=3$$  

$$(х+5)^2+у^2=3$$