Question

Determinar o comprimento da corda que a circunferência x2 + y2 - 4x + 6y - 12 =0 determina sobre a reta (r) 3x - 4y - 3 = 0.

Answer 1

$\mathrm{x^2+y^2-4x+6y-12=0}\\ \mathrm{x^2-4x+(y^2+6y)=12}\\ \mathrm{(y^2+6y+9)=12+9-x^2+4x}\\ \mathrm{(y+3)^2=21-x^2+4x}\\ \mathrm{y+3=\sqrt{21-x^2+4x}}\\ \mathrm{f(x)=\sqrt{21-x^2+4x}-3}\\\\ \mathrm{3x-4y-3=0\ \to\ 4y=3x-3\ \to\ g(x)=\dfrac{3x}{4}-\dfrac{3}{4}}$

$\textbf{Abscissas de intersec\c{c}\~ao:}\\\\ \mathrm{f(x)=g(x)\ \to\ \sqrt{21-x^2+4x}-3=\dfrac{3x}{4}-\dfrac{3}{4}}\\ \mathrm{\sqrt{21-x^2+4x}=\dfrac{3x-3}{4}+\dfrac{12}{4}}\\ \mathrm{\big(\sqrt{21-x^2+4x}\big)^2=\bigg(\dfrac{3x+9}{4}\bigg)^2}\\ \mathrm{21-x^2+4x=\dfrac{9x^2+54x+81}{16}}\\\\ \mathrm{336-16x^2+64x=9x^2+54x+81}\\ \mathrm{-16x^2-9x^2+64x-54x+336-81=0}\\ \mathrm{-25x^2+10x+255=0\ \to\ 5x^2-2x-51=0}$

$\mathrm{x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4.5.(-51)}}{2.5}}\\\\ \mathrm{x=\dfrac{2\pm\sqrt{4+1020}}{10}=\dfrac{2\pm\sqrt{1024}}{10}}\\\\ \mathrm{x=\dfrac{2\pm32}{10}=\dfrac{1\pm16}{5}}\\\\ \mathrm{x_1=\dfrac{1+16}{5}=\dfrac{17}{5}\ \ \|\ \ x_2=\dfrac{1-16}{5}=\dfrac{-15}{5}=-3}$

$\textbf{Ordenadas de intersec\c{c}\~ao:}\\\\ \mathrm{y=\dfrac{3x}{4}-\dfrac{3}{4}\ \to\ y_1=\dfrac{3.\frac{17}{5}}{4}-\dfrac{3}{4}=\dfrac{3(\frac{17}{5}-1)}{4}=}\\\\ \mathrm{=\dfrac{3(\frac{17}{5}-\frac{5}{5})}{4}=\dfrac{3(\frac{12}{5})}{4}=\dfrac{36}{5}.\dfrac{1}{4}\ \to\ y_1=\dfrac{9}{5}}\\\\ \mathrm{y_2=\dfrac{3.(-3)}{4}-\dfrac{3}{4}=\dfrac{-9}{4}-\dfrac{3}{4}=\dfrac{-12}{4}\ \to\ y_2=-3}$

$\textbf{Coordenadas de intersec\c{c}\~ao:}\\\\ \mathrm{f(x)\cap g(x)=\{A(x_1,y_1);B(x_2,y_2)\}}\\\\ \mathrm{f(x)\cap g(x)=\bigg\{A\bigg(\dfrac{17}{5},\dfrac{9}{5}\bigg);B\big(-3,-3\big)\bigg\}}$

$\textbf{O comprimento da corda ser\'a igual}\\ \textbf{dist\^ancia entre esses dois pontos:}\\\\ \mathrm{d_{AB}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}=}\\\\ \mathrm{=\sqrt{\bigg(-3-\dfrac{17}{5}\bigg)^2+\bigg(-3-\dfrac{9}{5}\bigg)^2}=}\\\\ \mathrm{=\sqrt{\bigg(\dfrac{-15-17}{5}\bigg)^2+\bigg(\dfrac{-15-9}{5}\bigg)^2}=}\\\\ \mathrm{=\sqrt{\bigg(\dfrac{-32}{5}\bigg)^2+\bigg(\dfrac{-24}{5}\bigg)^2}=\sqrt{\dfrac{1024}{25}+\dfrac{576}{25}}=}\\\\ \mathrm{=\sqrt{\dfrac{1600}{25}}=\dfrac{40}{5}=8}$

$\boxed{\mathbf{c_{corda}=d_{AB}=8\ u.c.}}$