As circunferências x² + y² + 4x + 2y - 20 = 0 e x² + y² - 8x - 6y - 11 = 0 são:
(A) secantes.
(B) tangentes internas.
(C) tangentes externas.
(D) externas.
(E) concêntricas.
(A) secantes.
(B) tangentes internas.
(C) tangentes externas.
(D) externas.
(E) concêntricas.
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✅ Após resolver os cálculos, concluímos que as referidas circunferências são:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf Secantes\:\:\:}}\end{gathered}$}[/tex]
Portanto, a opção correta é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf Alternativa\: A\:\:\:}}\end{gathered}$}[/tex]
Sejam as equações das circunferências:
[tex]\Large\begin{cases} \lambda: x^{2} + y^{2} + 4x + 2y - 20 = 0\\\gamma: x^{2} + y^{2} - 8x - 6y - 11 = 0\end{cases}[/tex]
Para resolver esta questão devemos encontrar os centros e os raios de ambas circunferências e depois, comparar a distância entre seus centros. Então, temos:
[tex]\Large\displaystyle\text{$\begin{gathered} X_{\lambda} = -\frac{D}{2A} = -\frac{4}{2\cdot1} = -2\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} Y_{\lambda} = -\frac{E}{2A} = -\frac{2}{2\cdot1} = -1\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:C_{\lambda} (-2, \, -1)\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} r_{\lambda} = \sqrt{\frac{D^{2} + E^{2} - 4AF}{4A^{2}}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{\frac{4^{2} + 2^{2} - 4\cdot1\cdot(-20)}{4\cdot1^{2}}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{\frac{16 + 4 + 80}{4}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 5\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:r_{\lambda} = 5\:u\cdot c\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} X_{\gamma} = -\frac{D}{2A} = -\frac{(-8)}{2\cdot1} = 4\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} Y_{\gamma} = -\frac{E}{2A} = -\frac{(-6)}{2\cdot1} = 3\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:C_{\gamma} (4, \, 3)\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} r_{\gamma} = \sqrt{\frac{D^{2} + E^{2} - 4AF}{4A^{2}}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{\frac{(-8)^{2} + (-6)^{2} - 4\cdot1\cdot(-11)}{4\cdot1^{2}}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{\frac{64 + 36 + 44}{4}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 6\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:r_{\gamma} = 6\:u\cdot c\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} d_{C_{\lambda}C_{\gamma}} = \sqrt{(X_{\gamma} - X_{\lambda})^{2} + (Y_{\gamma} - Y_{\lambda})^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{(4 - (-2))^{2} + (3 - (-1))^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{(4 + 2)^{2} + (3 + 1)^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{6^{2} + 4^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{36 + 16}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{52}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 2\sqrt{13}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:d_{C_{\lambda}C_{\gamma}} = 2\sqrt{13}\cong7,211\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} r_{\lambda} + r_{\gamma} = 5 + 6 = 11\:u\cdot c\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} |r_{\lambda} - r_{\gamma}| = |5 - 6| = |-1| = 1\:u\cdot c \end{gathered}$}[/tex]
Sendo:
[tex]\Large\displaystyle\text{$\begin{gathered} |r_{\lambda} - r_{\gamma}| < d_{C_{\lambda}C_{\gamma}} < r_{\lambda} + r_{\gamma}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} 1 < d_{C_{\lambda}C_{\gamma}} < 11 \end{gathered}$}[/tex]
✅ Então as circunferências são:
[tex]\Large\displaystyle\text{$\begin{gathered} \textrm{Secantes}\end{gathered}$}[/tex]
Saiba mais:
Veja a solução gráfica representada na figura: