Question

З'ясуйте взаємно розташування двох кіл(дотик,перетин або немає спільних точок) (х-1)^2+(у-2)^2=4 і (х-6)^2+(у-2)^2=9 ​

Answer 1

$\left \{ {{(x-1)^2+(y-2)^2=4} \atop {(x-6)^2+(y-2)^2=9}} \right. =\\\\= \left \{ {{(x-1)^2+(y-2)^2=4} \atop {(x-6)^2-(x-1)^2=9-4}} \right. =\\\\= \left \{ {{(x-1)^2+(y-2)^2=4} \atop {x^2-12x+36-(x^2-2x+1)=5}} \right. =\\\\= \left \{ {{(x-1)^2+(y-2)^2=4} \atop {x^2-12x+36-x^2+2x-1=5}} \right. =\\\\= \left \{ {{(x-1)^2+(y-2)^2=4} \atop {-10x+35=5}} \right. =\\\\= \left \{ {{(x-1)^2+(y-2)^2=4} \atop {-10x=-30}} \right. =\\\\= \left \{ {{(x-1)^2+(y-2)^2=4} \atop {x=3}} \right. =$

$= \left \{ {{(x-1)^2+(y-2)^2=4} \atop{x=3}} \right. =\\\\= \left \{ {{(3-1)^2+(y-2)^2=4} \atop {x=3}} \right. =\\\\= \left \{ {{4+y^2-2y+4=4} \atop {x=3}} \right. =\\\\= \left \{ {{y^2-2y+4=0} \atop {x=3}} \right. =\\\\= \left \{ {{(y-2)(y-2)=0} \atop {x=3}} \right. =\\\\= \left \{ {{y=2} \atop {x=3}} \right..$

Два кола дотикаються у точці 3;2