Question

Determine a posição relativa entre cada reta e cada uma das circunferência
r: 3x - 4y = -10
s: 12x + 9y = - 15
t: 4x - 3y = - 35

$a) \lambda: (x - 4)^{2} + (y - 3)^{2} = 144$
$b) \tau: (x - 5) ^{2} + (y - 1)^{2} = 36$
$c) \phi :(x + 3)^{2} + (y + 6)^{2} = 9$
​

Answer 1

Resposta:

a)

Centro =(4,3) raio =12

r: 3x - 4y = -10

d=|3*4-4*3+10|/√(3²+4²) = 10/5 =2  < 12  é secante

s: 12x + 9y = - 15

d=|12*4+9*3+15|/√(12²+9²) = 90/15 =6  < 12  é secante

t: 4x - 3y = - 35

d=|4*4 - 3*3+35|/√(4²+3²) =42/5=8,4 < 12 secante

b)

Centro (5,1)  raio =6

r: 3x - 4y = -10

d=|3*5-4*1+10|/√(3²+4²) = 21/5 =4,2  <6  secante

s: 12x + 9y = - 15

d=|12*5+9*1+15|/√(12²+9²) = 81/15 =5,4 < 6  é secante

t: 4x - 3y = - 35

d=|4*5 - 3*1+35|/√(4²+3²) =52/5=10,4 > 6  externa

c)

Centro (-3,-6)  raio =3

r: 3x - 4y = -10

d=|3*(-3)-4*(-6)+10|/√(3²+4²) = 25/5 =5  >3 externa

s: 12x + 9y = - 15

d=|12*(-3)+9*(-6)+15|/√(12²+9²) = 75/15 =5 >3 externa

t: 4x - 3y = - 35

d=|4*(-3) - 3*(-6)+35|/√(4²+3²) =41/5=8,2 > 3  externa

Answer 2

r: 3x - 4y = -10

s: 12x + 9y = - 15

t: 4x - 3y = - 35

$a) \lambda: (x - 4)^{2} + (y - 3)^{2} = 144$

$C(4,3)\\R=\sqrt{144}=12$

$\textcolor{dodgerblue}{d_{C,r}=\frac{ax_{p} </p><p>+by_{p}+c}{\sqrt{{a}^{2}+{b}^{2}}}}$

$\textcolor{dodgerblue}{d_{C,r}=\frac{3.4</p><p>-4.3+10}{\sqrt{{3}^{2}+{(-4)}^{2}}}}$

$\textcolor{dodgerblue}{d_{C,r}=\frac{10}{\sqrt{9+16}}}$

$\textcolor{dodgerblue}{d_{C,r}=\frac{10}{\sqrt{25}}}$

$\textcolor{dodgerblue}{d_{C,r}=\frac{10}{5}}$

$\textcolor{dodgerblue}{d_{C,r}=2<12}$

$\textcolor{dodgerblue}{\mathfrak{r\:é\:secante\:a\:\lambda}}$

$b) \tau: (x - 5) ^{2} + (y - 1)^{2} = 36$

$C(5,1)\\R=\sqrt{36}=6$

$\textcolor{dodgerblue}{d_{C,r}=\frac{ax_{p} </p><p>+by_{p}+c}{\sqrt{{a}^{2}+{b}^{2}}}}$

$\textcolor{dodgerblue}{d_{C,r}=\frac{12.5+9.1+15}{\sqrt{{12}^{2}+{9}^{2}}}}$

$\textcolor{dodgerblue}{d_{C,r}=\frac{60+9+15}{\sqrt{144+81}}}$

$\textcolor{dodgerblue}{d_{C,r}=\frac{84}{\sqrt{225}}}$

$\textcolor{dodgerblue}{d_{C,r}=\frac{84}{15}<6}$

$\textcolor{dodgerblue}{\mathfrak{s\:é\:secante\:a\:\tau}}$

$c) \phi :(x + 3)^{2} + (y + 6)^{2} = 9$

Resumo da ópera:

$d_{Centro,reta}<raio$

a reta é secante a circunferência

$d_{Centro,reta}=raio$ a reta é tangente a circunferência.

$d_{Centro,reta}>raio$

A reta é externa a circunferência