معادلة الدائرة

$\left(x+1\right)^2\ +\left(y+1\right)^2\ =\ 3$

$\left(x+1\right)^2\ +\ \left(y-\ 1\ \right)^2\ =\ 3$

$\left(x+1\right)^2+\ \left(y+1\right)^2=9$

$\left(x\ -\ 1\ \right)^2+\left(y-1\right)^2=9$

$\left(x\ -\ 7\right)^2+\ y^2=9$

$x^2+\left(y-7\right)^2=9$

$\left(x-7\right)^2+y^2=3$

$x^2+\left(y-7\right)^2=3$

2

3

4

16

$\left(x+1\right)^2+\left(y-3\right)^2=16$

$\left(x-1\right)^2+\left(y+3\right)^2=16$

$\left(x-1\right)^2+\left(y+3\right)^2=4$

$\left(x+1\right)^2+\left(y-3\right)^2=4$

$\left(x+1\right)+\left(y+1\right)=9$

x+y=9

$x^2+y^2=3$

$x^2+y^2=9$