Me ajudem, por favor
A equação geral da circunferência com centro C(4, - 1) e tangente a reta x + y + 5 = 0 é:
(A) x² + y² - 8x + 2y + 17 = 0.
(B) x² + y² - 8x + 2y + 9 = 0.
(C) x² + y² - 8x + 2y - 15 = 0.
(D) x² + y² - 8x + 2y - 7 = 0.
(E) x² + y² - 8x + 2y = 0.
A equação geral da circunferência com centro C(4, - 1) e tangente a reta x + y + 5 = 0 é:
(A) x² + y² - 8x + 2y + 17 = 0.
(B) x² + y² - 8x + 2y + 9 = 0.
(C) x² + y² - 8x + 2y - 15 = 0.
(D) x² + y² - 8x + 2y - 7 = 0.
(E) x² + y² - 8x + 2y = 0.
More Questions From This User See All
Lista de comentários
✅ Após resolver os cálculos, concluímos que a equação geral da circunferência tangente à reta "s" é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf \lambda: x^{2} + y^{2} - 8x + 2y - 15 = 0\:\:\:}}\end{gathered}$}[/tex]
Portanto, a opção correta é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf Alternativa\:C\:\:\:}}\end{gathered}$}[/tex]
Sejam os dados:
[tex]\Large\begin{cases} C(4, -1)\\s: x + y + 5 = 0\end{cases}[/tex]
Para resolver esta questão devemos:
[tex]\Large\displaystyle\text{$\begin{gathered} r = d_{\overline{Cs}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{|ax_{c} + by_{c} + c|}{\sqrt{a^{2} + b^{2}}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{|1\cdot4 + 1\cdot(-1) + 5|}{\sqrt{1^{2} + 1^{2}}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{|4 - 1 + 5|}{\sqrt{1 + 1}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{|8|}{\sqrt{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{8}{\sqrt{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{8}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{8\sqrt{2}}{(\sqrt{2})^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{8\sqrt{2}}{2}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 4\sqrt{2}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\: r = 4\sqrt{2}\:u\cdot c\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} (x - x_{c})^{2} + (y - y_{c})^{2} = r^{2}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} (x - 4)^{2} + (y - (-1))^{2} = (4\sqrt{2})^{2}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} (x - 4)^{2} + (y + 1)^{2} = 4^{2}\cdot(\sqrt{2})^{2}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} (x - 4)^{2} + (y + 1)^{2} = 32\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} x^{2} - 8x + 16 + y^{2} + 2y + 1 = 32\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} x^{2} + y^{2} - 8x + 2y + 16 + 1 - 32 = 0\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} x^{2} + y^{2} - 8x + 2y - 15 = 0\end{gathered}$}[/tex]
✅ Portanto, a equação procurada é:
[tex]\Large\displaystyle\text{$\begin{gathered} \lambda: x^{2} + y^{2} - 8x + 2y - 15 = 0\end{gathered}$}[/tex]
Saiba mais:
Veja a solução gráfica representada na figura: