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
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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