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

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

 

$\text{a) }$   $S=\left(-1,3\right)$ , $r=3$  

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

${\left(x+1\right)}^2+{\left(y-3\right)}^2=9$

---

$\text{b) }$   $S=\left(0,0\right)$ , $r=11$  

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

$x^2+y^2=121$

---

$\text{c) }$   $S=\left(-2,-5\right)$ , $r=1$  

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

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

---

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

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

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

---

$\text{e) }$   $S=\left(6,-4\right)$ , $r=5$  

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

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

---

$\text{f) }$   $S=\left(-\frac{1}{3},0\right)$ , $r=\sqrt{3}$  

${\left(x-\left(-\frac{1}{3}\right)\right)}^2+{\left(y-0\right)}^2={\sqrt{3}}^2$

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