Question

seja ʎ a circunferência que passa pelos pontos P (5,6),Q (3,2) e R(7,4) e S a reta tangente à ʎ que passa por (0,-2).determine a distância do ponto T(-2,5) à reta S​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf P(5,6) \Leftrightarrow Q(3,2) \Leftrightarrow R(7,4)$

$\sf d_{PC} = d_{QC} = d_{RC} = r$

$\sf (x_P - x_C)^2 + (y_P - y_C)^2 = (x_Q - x_C)^2 + (y_Q - y_C)^2$

$\sf (5 - x_C)^2 + (6 - y_C)^2 = (3 - x_C)^2 + (2 - y_C)^2$

$\sf [\:25 - 10x_C + (x_C)^2\:] + [\:36 - 12y_C + (y_C)^2\:] = [\:9 - 6x_C + (x_C)^2\:] + [\:4 - 4y_C + (y_C)^2\:]$

$\sf [\:25 - 10x_C\:] + [\:36 - 12y_C] = [\:9 - 6x_C] + [\:4 - 4y_C\:]$

$\sf 61 - 10x_C - 12y_C = 13 - 6x_C - 4y_C$

$\sf - 4x_C - 8y_C = -48$

$\sf x_C + 2y_C = 12$

$\sf (x_Q - x_C)^2 + (y_Q - y_C)^2 = (x_R - x_C)^2 + (y_R - y_C)^2$

$\sf (3 - x_C)^2 + (2 - y_C)^2 = (7 - x_C)^2 + (4 - y_C)^2$

$\sf [\:9 - 6x_C + (x_C)^2\:] + [\:4 - 4y_C + (y_C)^2\:] = [\:49 - 14x_C + (x_C)^2\:] + [\:16 - 8y_C + (y_C)^2\:]$

$\sf [\:9 - 6x_C] + [\:4 - 4y_C] = [\:49 - 14x_C] + [\:16 - 8y_C]$

$\sf 13 - 6x_C - 4y_C = 65 - 14x_C - 8y_C$

$\sf 8x_C + 4y_C = 52$

$\sf 2x_C + y_C = 13$

$\begin{cases}\sf x_C + 2y_C = 12\\\sf 2x_C + y_C = 13\end{cases}$

$\begin{cases}\sf x_C + 2y_C = 12\\\sf -4x_C - 2y_C = -26\end{cases}$

$\sf -3x_C = -14\rightarrow x_C = \dfrac{14}{3}$

$\sf \dfrac{14}{3} + 2y_C = 12\rightarrow y_C = \dfrac{11}{3}$

$\sf C\left[\dfrac{14}{3},\dfrac{11}{3}\right]$

$\sf r^2 = (x_P - x_C)^2 + (y_P - y_C)^2$

$\sf r^2 = \left(5 - \dfrac{14}{3}\right)^2 + \left(6 - \dfrac{11}{3}\right)^2$

$\sf r^2 = \left(\dfrac{1}{3}\right)^2 + \left(\dfrac{7}{3}\right)^2$

$\sf r^2 = \dfrac{1}{9} + \dfrac{49}{9}$

$\sf r^2 = \dfrac{50}{9}$

$\sf \left(x - \dfrac{14}{3}\right)^2 + \left(y - \dfrac{11}{3}\right)^2 = \dfrac{50}{9}$

$\sf \left(x^2 - \dfrac{28}{3}\:x + \dfrac{196}{9}\right) + \left(y^2 - \dfrac{22}{3}\:y + \dfrac{121}{9}\right) = \dfrac{50}{9}$

$\sf 9x^2 - 84x + 196 + 9y^2 - 66y + 121 - 50 = 0$

$\boxed{\sf \lambda:9x^2 + 9y^2 - 84x - 66y + 267 = 0}$

$\sf U(0,-2) \Leftrightarrow T(-2,5)$

$\sf M_{UC} = \left\{\left(0 + \dfrac{14}{3}\right)/2,\left(-2 + \dfrac{11}{3}\right)/2\right\}$

$\boxed{\sf M_{UC} = \left\{\dfrac{7}{3},\dfrac{5}{6}\right\}}\rightarrow\textsf{centro da circunfer{\^e}ncia de apoio}$

$\sf r = d_{MC}$

$\sf r^2 = (x_M - x_C)^2 + (y_M - y_C)^2$

$\sf r^2 = \left(\dfrac{7}{3} - \dfrac{14}{3}\right)^2 + \left(\dfrac{5}{6} - \dfrac{11}{3}\right)^2$

$\sf r^2 = \left(- \dfrac{7}{3}\right)^2 + \left( - \dfrac{17}{6}\right)^2$

$\sf r^2 = \dfrac{49}{9} + \dfrac{289}{36}$

$\sf r^2 = \dfrac{485}{36}$

$\sf \left(x - \dfrac{7}{3}\right)^2 + \left(y - \dfrac{5}{6}\right)^2 = \dfrac{485}{36}$

$\sf \left(x^2 -\dfrac{14}{3}\:x + \dfrac{49}{9}\right) + \left(y^2 - \dfrac{5}{3}\:y + \dfrac{25}{36}\right) = \dfrac{485}{36}$

$\sf 36x^2 - 168x + 196 + 36y^2 - 60y + 25 - 485 = 0$

$\boxed{\sf \gamma:36x^2 + 36y^2 - 168x - 60y -264 = 0}$

$\begin{cases}\sf 3x^2 + 3y^2 - 28x - 22y + 89 = 0\\\sf 6x^2 + 6y^2 - 28x - 10y -44 = 0\end{cases}$

$\begin{cases}\sf -6x^2 - 6y^2 + 56x + 44y - 178 = 0\\\sf 6x^2 + 6y^2 - 28x - 10y -44 = 0\end{cases}$

$\sf 14x + 17y = 111$

$\sf 3\left[\dfrac{111 - 17y}{14}\right]^2 + 3y^2 - 28\left[\dfrac{111 - 17y}{14}\right] - 22y + 89 = 0$

$\sf 3\left[\dfrac{12.321 - 3.774y + 289y^2}{196}\right] + 3y^2 - 222 + 34y - 22y + 89 = 0$

$\sf \dfrac{36.963 - 11.322y + 867y^2}{196} + 3y^2 - 222 + 34y - 22y + 89 = 0$

$\sf 36.963 - 11.322y + 867y^2 + 588y^2 - 43.512 + 6.664y - 4.312y + 17.444 = 0$

$\sf 1.455y^2 - 8.970y + 10.895 = 0$

$\sf 291y^2 - 1.794y + 2.179 = 0$

$\sf a = 291 \Leftrightarrow b = -1.794 \Leftrightarrow c = 2.179$

$\sf \Delta = b^2 - 4.a.c$

$\sf \Delta = (-1.794)^2 - 4.291.2179$

$\sf \Delta = 3.218.436 - 2.536.356$

$\sf \Delta = 682.080$

$\sf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{1794 \pm \sqrt{682080}}{582} \rightarrow \begin{cases}\sf{y' = \dfrac{1794 + 28\sqrt{870}}{582} = 4.52}\\\\\sf{y'' = \dfrac{1794 - 28\sqrt{870}}{582} = 1.66}\end{cases}}$

$\boxed{\sf A(2.46,4.52) \Leftrightarrow B(5.91,1.66)}\rightarrow\textsf{pontos onde as retas tangenciam a circunfer{\^e}ncia}$

$\sf m = \dfrac{\Delta_Y}{\Delta_X} = \dfrac{y_T - y_A}{x_T - x_A} = \dfrac{-2-4.52}{0-2.46} = \dfrac{6.52}{2.46} = 2.65$

$\sf y - y_0 = m(x - x_0)$

$\sf y + 2 = 2.65(x - 0)$

$\boxed{\sf s:y = 2.65x - 2}\rightarrow\textsf{uma das retas que tangencia a circunfer{\^e}ncia }$

$\sf d_{T,s} = |\:\dfrac{ax_0 + by_0 + c}{\sqrt{a^2 + b^2}}\:|$

$\sf d_{T,s} = |\:\dfrac{(2.65)(-2) + (-1)(5) - 2}{\sqrt{(2.65)^2 + (-1)^2}}\:|$

$\sf d_{T,s} = |\:\dfrac{-5.3 - 5 - 2}{\sqrt{7.02 + 1}}\:|$

$\sf d_{T,s} = |\:\dfrac{-12.3}{2.83}\:|$

$\boxed{\boxed{\sf d_{T,s} = 4.34}}\leftarrow\textsf{primeira solu}\sf c_{\!\!,}\textsf{{\~a}o poss{\'i}vel}$

$\sf m = \dfrac{\Delta_Y}{\Delta_X} = \dfrac{y_T - y_A}{x_T - x_A} = \dfrac{-2-1.66}{0-5.91} = \dfrac{3.66}{5.91} = 0.62$

$\sf y - y_0 = m(x - x_0)$

$\sf y + 2 = 0.62(x - 0)$

$\boxed{\sf r:y = 0.62x - 2}\rightarrow\textsf{outra reta que tangencia a circunfer{\^e}ncia }$

$\sf d_{T,r} = |\:\dfrac{ax_0 + by_0 + c}{\sqrt{a^2 + b^2}}\:|$

$\sf d_{T,r} = |\:\dfrac{(0.62)(-2) + (-1)(5) - 2}{\sqrt{(0.62)^2 + (-1)^2}}\:|$

$\sf d_{T,r} = |\:\dfrac{-1.24 - 5 - 2}{\sqrt{0.38 + 1}}\:|$

$\sf d_{T,r} = |\:\dfrac{-8.24}{1.17}\:|$

$\boxed{\boxed{\sf d_{T,r} = 7.04}}\leftarrow\textsf{segunda solu}\sf c_{\!\!,}\textsf{{\~a}o poss{\'i}vel}$