$\text{a)}$  

$\{\begin{array}{l}{\left(x+2\right)}^2+{\left(y-3\right)}^2=25\\ x-2y+18=0\end{array}$  

$\{\begin{array}{l}{\left(x+2\right)}^2+{\left(y-3\right)}^2=25\\ x=2y-18\end{array}$  

$\{\begin{array}{l}{\left(2y-18+2\right)}^2+{\left(y-3\right)}^2=25\\ x=2y-18\end{array}$       

Rozwiązując pierwsze równanie:

${\left(2y-16\right)}^2+{\left(y-3\right)}^2=25$  

$4y^2-64y+256+y^2-6y+9-25=0$  

$5y^2-70y+240=0\mid :5$  

$y^2-14y+48=0$  

$\Delta ={\left(-14\right)}^2-4\cdot 1\cdot 48=196-192=4$  

$\Delta >0\Rightarrow \text{dwa punkty wspoˊlne}$  

---

$\text{b)}$  

$\{\begin{array}{l}{\left(x+2\right)}^2+{\left(y-3\right)}^2=25\\ 2x-y-6=0\end{array}$  

$\{\begin{array}{l}{\left(x+2\right)}^2+{\left(y-3\right)}^2=25\\ 2x-6=y\end{array}$  

$\{\begin{array}{l}{\left(x+2\right)}^2+{\left(2x-6-3\right)}^2=25\\ 2x-6=y\end{array}$  

Rozwiązując pierwsze równanie:

${\left(x+2\right)}^2+{\left(2x-9\right)}^2=25$  

$x^2+4x+4+4x^2-36x+81-25=0$  

$5x^2-32x+60=0$  

$\Delta ={\left(-32\right)}^2-4\cdot 5\cdot 60=1024-1200<0$  

$\Delta <0\Rightarrow \text{brak punktoˊw wspoˊlnych}$  

---

$\text{c)}$  

$\{\begin{array}{l}{\left(x+2\right)}^2+{\left(y-3\right)}^2=25\\ 4x-3y-8=0\end{array}$  

$\{\begin{array}{l}{\left(x+2\right)}^2+{\left(y-3\right)}^2=25\\ 4x=3y+8\end{array}$

$\{\begin{array}{l}{\left(x+2\right)}^2+{\left(y-3\right)}^2=25\\ x=\frac{3}{4}y+2\end{array}$

$\{\begin{array}{l}{\left(\frac{3}{4}y+2+2\right)}^2+{\left(y-3\right)}^2=25\\ x=\frac{3}{4}y+2\end{array}$    

Rozwiązując pierwsze równanie:

${\left(\frac{3}{4}y+4\right)}^2+{\left(y-3\right)}^2=25$  

$\frac{9}{16}{y}^2+2\cdot \frac{3}{4}y\cdot 4+16+y^2-6y+9=25$  

$\frac{9}{16}{y}^2+6y+16+y^2-6y+9=25$  

$\frac{25}{16}{y}^2=0$  

$y^2=0$  

$y=0$  

Jeden punkt wspólny.