Qual dos pontos abaixo está interno a circunferência de equação x² + y² - 6x - 8y + 5 = 0?
(A) (2, 7).
(B) (1, 0).
(C) (3, -4).
(D) (0, 0).
(E) (5, 0)
(A) (2, 7).
(B) (1, 0).
(C) (3, -4).
(D) (0, 0).
(E) (5, 0)
More Questions From This User See All
Lista de comentários
Verified answer
✅ Após resolver os cálculos, concluímos que o único ponto interno à circunferência é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf A(2, 7)\:\:\:}}\end{gathered}$}[/tex]
Seja a equação da circunferência:
[tex]\Large\displaystyle\text{$\begin{gathered} \lambda: x^{2} + y^{2} - 6x - 8y + 5 = 0\end{gathered}$}[/tex]
Sabendo que equação da circunferência pode ser montada sobre a sua forma geral obedecendo à seguinte fórmula:
[tex]\Large\displaystyle\text{$\begin{gathered} ax^{2} + by^{2} + cxy + dx + ey + f = 0\end{gathered}$}[/tex]
Para resolver esta questão devemos:
[tex]\Large\displaystyle\text{$\begin{gathered} x_{O} = -\frac{d}{2a} = - \frac{(-6)}{2\cdot1} = 3\end{gathered}$}[/tex]
E...
[tex]\Large\displaystyle\text{$\begin{gathered} y_{O} = -\frac{e}{2a} = - \frac{(-8)}{2\cdot1} = 4\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:O(3, 4)\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} r = \sqrt{\frac{d^{2} + e^{2} - 4af}{4a^{2}}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{\frac{(-6)^{2} + (-8)^{2} - 4\cdot1\cdot5}{4\cdot1^{2}}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{\frac{36 + 64 - 20}{4}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{\frac{80}{4}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{20}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 2\sqrt{5}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:r = 2\sqrt{5}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d(AO) = \sqrt{(X_{O} - X_{A})^{2} + (Y_{O} - Y_{A})^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{(3 - 2)^{2} + (4 - 7)^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{10}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d(AO) < r\:\:\therefore\:\:\:A\:\acute{e}\:Interno\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d(BO) = \sqrt{(X_{O} - X_{B})^{2} + (Y_{O} - Y_{B})^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{(3 - 1)^{2} + (4 - 0)^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 2\sqrt{5}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d(BO) = r\:\:\therefore\:\:\:B\in\lambda\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d(CO) = \sqrt{(X_{O} - X_{C})^{2} + (Y_{O} - Y_{C})^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{(3 - 3)^{2} + (4 - (-4))^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 8\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d(CO) > r\:\:\therefore\:\:\:C\:\acute{e}\:Externo\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d(DO) = \sqrt{(X_{O} - X_{D})^{2} + (Y_{O} - Y_{D})^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{(3 - 0)^{2} + (4 - 0)^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 5\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d(DO) > r\:\:\therefore\:\:\:D\:\acute{e}\:Externo\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d(EO) = \sqrt{(X_{O} - X_{E})^{2} + (Y_{O} - Y_{E})^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{(3 - 5)^{2} + (4 - 0)^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 2\sqrt{5}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d(EO) = r\:\:\therefore\:\:\:E\in\lambda\end{gathered}$}[/tex]
✅ Portanto, o único ponto interno é:
[tex]\Large\displaystyle\text{$\begin{gathered} A(2, 7)\end{gathered}$}[/tex]
Saiba mais:
Veja a solução gráfica da questão representada na figura: