Question

A reta x-y=0 estabelece na circunferência x²+y²-4x-2y+p=0 uma corda de comprimento √14. Calcule o valor de p.

Answer 1

Resposta:  p = 1.

Explicação passo-a-passo:

Os pontos da reta x − y = 0 são da forma P(x, x), ou seja, a abscissa é igual à ordenada.

Sejam A(x₁, x₁) e B(x₂, x₂) as extremidades da corda. De acordo com o enunciado, teremos

    $\mathsf{\|AB\|=\sqrt{14}}\\\\ \mathsf{\sqrt{(x_2-x_1)^2+(x_2-x_1)^2}=\sqrt{14}}\\\\ \mathsf{\sqrt{2\cdot (x_2-x_1)^2}=\sqrt{14}}\\\\ \mathsf{\sqrt{2}\cdot |x_2-x_1|=\sqrt{14}}\\\\ \mathsf{|x_2-x_1|=\dfrac{\sqrt{14}}{\sqrt{2}}}\\\\\\ \mathsf{|x_2-x_1|=\sqrt{\dfrac{14}{2}}}\\\\\\ \mathsf{|x_2-x_1|=\sqrt{7}\qquad (i)}$

Por outro lado, A e B são os pontos de interseção da reta x − y = 0 com a circunferência de equação x² + y² − 4x − 2y + p = 0.

Fazendo y = x na equação da circunferência, obtemos

    $\mathsf{x^2+x^2-4x-2x+p=0}\\\\ \mathsf{2x^2-6x+p=0}\\\\\\ \mathsf{x^2-3x+\dfrac{p}{2}=0\quad\longrightarrow\quad a=1,~b=-3,~c=\dfrac{p}{2}}\\\\\\\\ \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(-3)^2-4\cdot 1\cdot \dfrac{p}{2}}\\\\\\ \mathsf{\Delta=9-2p}$

com $\mathsf{p<\dfrac{9}{2}.}$

Então, encontramos

    $\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{x=\dfrac{-(-3)\pm\sqrt{9-2p}}{2\cdot 1}}\\\\\\ \mathsf{x=\dfrac{3\pm\sqrt{9-2p}}{2}}\\\\\\ \mathsf{x_1=\dfrac{3-\sqrt{9-2p}}{2}\quad e\quad x_2=\dfrac{3+\sqrt{9-2p}}{2}}$

Substituindo em (i), devemos ter

    $\mathsf{\left|\dfrac{3+\sqrt{9-2p}}{2}-\dfrac{3-\sqrt{9-2p}}{2}\right|=\sqrt{7}}\\\\\\ \mathsf{\left|\dfrac{\diagup\!\!\!\! 3+\sqrt{9-2p}-\diagup\!\!\!\! 3+\sqrt{9-2p}}{2}\right|=\sqrt{7}}\\\\\\ \mathsf{\left|\dfrac{\diagup\!\!\!\! 2\sqrt{9-2p}}{\diagup\!\!\!\! 2}\right|=\sqrt{7}}\\\\\\ \mathsf{\left|\sqrt{9-2p}\right|=\sqrt{7}}\\\\\\ \mathsf{\sqrt{9-2p}=\sqrt{7}}\\\\ \mathsf{9-2p=7}$

    $\mathsf{2p=9-7}\\\\ \mathsf{2p=2}$

    $\mathsf{p=1\quad\longleftarrow\quad resposta.}$

Bons estudos! :-)