Question

A reta de equação x+2y=3 intercepta a circunferência x^2+y^2=4 nos pontos AeB.Considerando as medidas em centímetros, o comprimento da corda AB é: a) 2/3 raiz de 5. b) 5/2raiz de 55. c) 6raiz de11 d) 3 raiz de 11 e) 2/5 raiz de55.

Answer 1

Os pontos de interseção pertencem à reta e à circunferência, então satisfazem as duas equações.

Isolando x na equação da reta:

$x+2y=3~~~\therefore~~~\boxed{\boxed{x=3-2y}}$

Então, um ponto qualquer da reta é (x,y) = (3 - 2y, y)

Os pontos de interseção com a circunferência terão essa forma, então, vamos substituir x por (3 - 2y) na equação da circunferência:

$x^{2}+y^{2}=4\\\\(3-2y)^{2}+y^{2}=4\\\\3^{2}-2\cdot3\cdot2y+(2y)^{2}+y^{2}=4\\\\9-12y+4y^{2}+y^{2}=4\\\\9-12y+5y^{2}-4=0\\\\5y^{2}-12y+5=0$

Resolvendo essa equação (achando as ordenadas dos pontos A e B):

$\Delta=b^{2}-4ac\\\Delta=(-12)^{2}-4\cdot5\cdot5\\\Delta=144-100\\\Delta=44\\\Delta=4\cdot11~~~\therefore~~~\sqrt{\Delta}=\sqrt{4}\sqrt{11}=2\sqrt{11}\\\\\\y=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-(-12)\pm2\sqrt{11}}{2\cdot5}=\dfrac{12\pm2\sqrt{11}}{2\cdot5}=\dfrac{6\pm\sqrt{11}}{5}$

Então, as soluções dessa equação são:

$\boxed{\boxed{y_{1}=\dfrac{6+\sqrt{11}}{5}}}\\\\\\\boxed{\boxed{y_{2}=\dfrac{6-\sqrt{11}}{5}}}$

Essas são as ordenadas dos pontos de interseção. Vamos achar as abscissas dos pontos:

$x=3-2y\\\\x_{1}=3-2y_{1}\\\\x_{1}=3-2\left(\dfrac{6+\sqrt{11}}{5}\right)=\dfrac{15}{5}+\dfrac{-12-2\sqrt{11}}{5}~~~\therefore~~~\boxed{\boxed{x_{1}=\dfrac{3-2\sqrt{11}}{5}}}$
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$x=3-2y\\\\x_{2}=3-2y_{2}\\\\x_{2}=3-2\left(\dfrac{6-\sqrt{11}}{5}\right)=\dfrac{15}{5}+\dfrac{-12+2\sqrt{11}}{5}~~~\therefore~~~\boxed{\boxed{x_{2}=\dfrac{3+2\sqrt{11}}{5}}}$

Então, os pontos A e B são:

$\left(\dfrac{6+\sqrt{11}}{5},\dfrac{3-2\sqrt{11}}{5}\right)~~~~e~~~~\left(\dfrac{6-\sqrt{11}}{5},\dfrac{3+2\sqrt{11}}{5}\right)$

O comprimento da corda AB é a distância entre esses pontos:

$l=d(A,B)\\\\l=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\\\\l=\sqrt{\left(\dfrac{3-2\sqrt{11}}{5}-\dfrac{3+2\sqrt{11}}{5}\right)^{2}+\left(\dfrac{6+\sqrt{11}}{5}-\dfrac{6-\sqrt{11}}{5}\right)^{2}}\\\\\\l=\sqrt{\left(-\dfrac{4\sqrt{11}}{5}\right)^{2}+\left(\dfrac{2\sqrt{11}}{5}\right)^{2}}\\\\\\l=\sqrt{\dfrac{(-4)^{2}(\sqrt{11})^{2}}{5^{2}}+\dfrac{2^{2}(\sqrt{11})^{2}}{5^{2}}}\\\\\\l=\sqrt{\dfrac{16\cdot11}{5}+\dfrac{4\cdot11}{25}}\\\\\\l=\sqrt{\dfrac{16\cdot11+4\cdot11}{25}}$

$l=\sqrt{\dfrac{20\cdot11}{25}}\\\\\\l=\sqrt{\dfrac{4\cdot11}{5}}\\\\\\l=\dfrac{\sqrt{4}}{\sqrt{5}}\sqrt{11}\\\\\\l=\dfrac{2}{\sqrt{5}}\sqrt{11}$

Racionalizando:

$l=\dfrac{2\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}}\sqrt{11}\\\\\\l=\dfrac{2}{5}\sqrt{5\cdot11}\\\\\\\boxed{\boxed{l=\dfrac{2}{5}\sqrt{55}}}$

Letra E