Odpowiedź :
Odpowiedź:
[tex](0,2),\ (3,5)[/tex]
Szczegółowe wyjaśnienie:
Należy rozwiązać układ równań.
[tex]\left \{ {{(x-2)^2+(y-3)^2=5} \atop {(x-5)^2+y^2=29}} \right. \\\left \{ {{x^2-4x+4+y^2-6y+9=5} \atop {x^2-10x+25+y^2=29}} \right. \\\left \{ {{x^2-4x+y^2-6y+8=0} \atop {x^2-10x+y^2-4=0}} \right. \\\left \{ {{x^2-4x+y^2-6y+8=0} \atop {y^2=-x^2+10x+4}} \right. \\\left \{ {{x^2-4x-x^2+10x+4-6y+8=0} \atop {y^2=-x^2+10x+4}} \right. \\\left \{ {{6x-6y+12=0\ |:6} \atop {y^2=-x^2+10x+4}} \right. \\\left \{ {{x-y+2=0} \atop {y^2=-x^2+10x+4}} \right.[/tex]
[tex]\left \{ {{y=x+2} \atop {(x+2)^2=-x^2+10x+4}} \right. \\\left \{ {{y=x+2} \atop {x^2+4x+4=-x^2+10x+4}} \right. \\\left \{ {{y=x+2} \atop {2x^2-6x=0\ |:2}} \right. \\\left \{ {{y=x+2} \atop {x^2-3x=0}} \right. \\\left \{ {{y=x+2} \atop {x(x-3)=0}} \right. \\\left \{ {{y=x+2} \atop {x=0}} \right. \vee \left \{ {{y=x+2} \atop {x=3}} \right.\\\left \{ {{y=2} \atop {x=0}} \right. \vee \left \{ {{y=5} \atop {x=3}} \right.[/tex]
Zatem szukane punkty to
[tex](0,2),\ (3,5)[/tex]