Question

1)dê o centro e o raio de cada circunferência
2)Dê a equação da circunferência de raio r e centro C, nos casos :
3) resolva, no universo dos números complexos, as equações :​

Answer 1

$\Large\boxed{\begin{array}{l}\boldsymbol{1}\\\rm a)~\sf (x-7)^2+(y-9)^2=36\\\sf C(7,9)\\\sf R^2=36\\\sf R=\sqrt{36}\\\sf R=6\\\rm b)~\sf x^2+y^2+4x-2y-11=0\\\sf x^2+4x+4+y^2-2y+1=11+4+1\\\sf (x+2)^2+(y-1)^2=16\\\sf C(-2,1)\\\sf R^2=16\\\sf R=\sqrt{16}\\\sf R=4\end{array}}$

$\Large\boxed{\begin{array}{l}\boldsymbol{2}\\\rm a)~\sf r=\sqrt{3}~~C(-3,6)\\\sf (x-[-3])^2+(y-6)^2=(\sqrt{3})^2\\\sf (x+3)^2+(y-6)^2=3\\\rm b)~\sf\bigg(x-\dfrac{1}{2}\bigg)^2+\bigg(y-\bigg[-\dfrac{1}{4}\bigg]\bigg)^2=\bigg(\dfrac{3}{4}\bigg)^2\\\\\sf\bigg(x-\dfrac{1}{2}\bigg)^2+\bigg(y+\dfrac{1}{4}\bigg)^2=\dfrac{9}{16}\end{array}}$

$\Large\boxed{\begin{array}{l}\boldsymbol{3}\\\rm a)~\sf x^2+4x=0\\\sf x\cdot(x+4)=0\\\sf x=0\\\sf x+4=0\\\sf x=-4\\\rm b)~\sf x^2+121=0\\\sf x^2=-121\\\sf x=\pm\sqrt{-121}\\\sf x=\pm11i\end{array}}$

$\Large\boxed{\begin{array}{l}\rm c)~\sf x^2-6x+13=0\\\sf\Delta= b^2-4ac\\\sf\Delta=(-6)^2-4\cdot1\cdot13\\\sf\Delta=36-52\\\sf\Delta=-16\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-6)\pm\sqrt{-16}}{2\cdot1}\\\\\sf x=\dfrac{6\pm4i}{2}\begin{cases}\sf x_1=\dfrac{6+4i}{2}=3+2i\\\\\sf x_2=\dfrac{6-4i}{2}=3-2i\end{cases}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm d)~\sf4x^2-4x+5=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-4)^2-4\cdot4\cdot5\\\sf\Delta=16-80\\\sf\Delta=-64\\\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-4)\pm\sqrt{-64}}{2\cdot4}\\\\\sf x=\dfrac{4\pm8i}{8}\begin{cases}\sf x_1=\dfrac{4+8i}{8}=\dfrac{\diagup\!\!\!\!4\cdot(1+2i)}{\diagup\!\!\!\!8}=\dfrac{1+2i}{2}\\\\\sf x_2=\dfrac{4+8i}{8}=\dfrac{\diagup\!\!\!\!4\cdot(1-2i)}{\diagup\!\!\!\!8}=\dfrac{1-2i}{2}\end{cases}\end{array}}$