Question

Considere as circunferências definidas por (x – 3)2 + (y – 2)2 = 16 e (x – 10)2 + (y – 2)2 = 9, representadas no mesmo plano cartesiano.

Answer 1

$Vamos~~resolver~~o sistema \\ \\ \left \{ {{(x-3)^2+(y-2)^2=16} \atop {(x-10)^2+(y-2)^2=9}} \right. \\ \\ isolar~~(y-2)^2~~em~~cada~~equac\tilde{a}o \\ \\ \left \{ {{(y-2)^2=16-(x-3)^2} \atop {(y-2)^2=9-(x-10)^2}} \right. \\ \\ igualando~~as~~equac\tilde{o}es \\ \\ 16-(x-3)^2=9-(x-10)^2 \\ \\ 16-(x^2-6x+9)=9-(x^2-20x+100) \\ \\ 16-x^2+6x-9=9-x^2+20x-100 \\ \\ -x^2+6x+7=-x^2+20x-91 \\ \\ -x^2+x^2+6x-20x=-91-7 \\ \\ -14x=-98~~~\times(-1) \\ \\ 14x=98 \\ \\ x=98\div14 \\ \\ x=7$
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$substituindo~~ x=7~~nas~~equac\tilde{o}es \\ \\ (x-3)^2+(y-2)^2=16 \\ (7-3)^2+(y-2)^2=16 \\ (4)^2+(y-2)^2=16 \\ 16+(y-2)^2=16 \\ (y-2)^2=16-16 \\ (y-2)^2=0 \\ y=2 \\ \\ em \\ \\ (x-10)^2+(y-2)^2=9 \\ (7-10)^2+(y-2)^2=9 \\ (-3)^2+(y-2)^2=9 \\ 9+(y-2)^2=9 \\ (y-2)^2=9-9 \\ (y-2)^2=0 \\ y=2$
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As coordenadas do ponto de interseção entre as circunferências (7,2)