Question

Uma circunferência do plano cartesiano tem equação (x - 2)² + (y - 5)² = 13. Obtenha os pontos de abscissa igual a 4 pertencentes a essa circunferência.

Answer 1

$\large\begin{array}{l} \textsf{Dada a circunfer\^encia de equa\c{c}\~ao}\\\\ \mathsf{(x-2)^2+(y-5)^2=13}\\\\ \textsf{queremos obter os pontos de abscissa 4.}\\\\ \textsf{Basta fazer }\mathsf{x=4}\textsf{ na equa\c{c}\~ao da circunfer\^encia, e resolver}\\ \textsf{a equa\c{c}\~ao obtida para y:}\\\\ \mathsf{(4-2)^2+(y-5)^2=13}\\\\ \mathsf{2^2+(y-5)^2=13}\\\\ \mathsf{4+(y-5)^2=13} \end{array}$

$\large\begin{array}{l} \mathsf{(y-5)^2=13-4}\\\\ \mathsf{(y-5)^2=9}\\\\ \mathsf{y-5=\pm \sqrt{9}}\\\\ \mathsf{y-5=\pm\,3}\\\\ \mathsf{y=\pm\,3+5}\\\\ \begin{array}{rcl} \mathsf{y=-3+5}&~\textsf{ ou }~&\mathsf{y=3+5}\\\\ \mathsf{y=2}&~\textsf{ ou }~&\mathsf{y=8} \end{array} \end{array}$


$\large\begin{array}{l} \textsf{Os pontos procurados s\~ao }\mathsf{(4,\,2)}\textsf{ e }\mathsf{(4,\,8).}\\\\\\ \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$