Równanie okręgu o środku w punkcie $S=\left(a,b\right)$  i promieniu $r>0$ :

${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$ .

 

$a)$   $S=\left(0,0\right)$ , $r=11$  

${\left(x-0\right)}^2+{\left(y-0\right)}^2={11}^2$

$x^2+y^2=121$

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$b)$   $S=\left(-2,-5\right)$ , $r=1$  

${\left(x-\left(-2\right)\right)}^2+{\left(y-\left(-5\right)\right)}^2=1^2$

${\left(x+2\right)}^2+{\left(y+5\right)}^2=1$

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$c)$   $S=\left(0,5\right)$ , $r=3\sqrt{2}$  

${\left(x-0\right)}^2+{\left(y-5\right)}^2={\left(3\sqrt{2}\right)}^2$

$x^2+{\left(y-5\right)}^2=18$

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$d)$   $S=\left(6,-4\right)$ , $r=5$  

${\left(x-6\right)}^2+{\left(y-\left(-4\right)\right)}^2=5^2$

${\left(x-6\right)}^2+{\left(y+4\right)}^2=25$