Tentukan persamaan garis singgung lingkaran (x-2)+(y-3)²=25 yang memiliki titik (9,2)
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Jawab:
3x - 4y - 19 = 0 dan 4x + 3y - 42 = 0
Penjelasan dengan langkah-langkah:
Tentukan kedudukan titik terhadap lingkaran. (9 - 2)² + (2 - 3)² = 50. Karena 50 > r², titik di luar lingkaran.
Cara cepat
Persamaan garis singgung lingkaran (x - a)² + (y - b)² = r² yang melalui titik (x₁, y₁) di luar lingkaran adalah y - y₁ = m(x - x₁) dengan [tex]\displaystyle m=\frac{(x_1-a)(y_1-b)\pm r\sqrt{(x_1-a)^2+(y_1-b)^2-r^2}}{(x_1-a)^2-r^2}[/tex]
[tex]\displaystyle m=\frac{(9-2)(2-3)\pm 5\sqrt{(9-2)^2+(2-3)^2-25}}{(9-2)^2-25}\\m_1=\frac{3}{4}~\textrm{atau}~m_2=-\frac{4}{3}[/tex]
Persamaan nya
[tex]\displaystyle \begin{matrix}y-y_1=m_1(x-x_1) & y-y_1=m_2(x-x_1)\\ y-2=\frac{3}{4}(x-9) & y-2=-\frac{4}{3}(x-9)\\4y-8=3x-27 & 3y-6=-4x+36\\3x-4y-19=0 & 4x+3y-42=0\end{matrix}[/tex]
Menggunakan rumus jarak titik ke garis
y - y₁ = m(x - x₁)
y - 2 = m(x - 9)
mx - y + 2 - 9m = 0 ← bentuk Ax + By + C = 0
[tex]\displaystyle r=\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}\\5=\frac{|m(2)+(-1)(3)+2-9m|}{\sqrt{m^2+(-1)^2}}[/tex]
(-7m - 1)² = 25(m² + 1)
49m² + 14m + 1 - 25m² - 25 = 0
24m² + 14m - 24 = 0
12m² + 7m - 12 = 0
(4m - 3)(3m + 4) = 0
m₁ = ¾ atau m₂ = -4/3
Persamaan nya
[tex]\displaystyle \begin{matrix}y-y_1=m_1(x-x_1) & y-y_1=m_2(x-x_1)\\ y-2=\frac{3}{4}(x-9) & y-2=-\frac{4}{3}(x-9)\\4y-8=3x-27 & 3y-6=-4x+36\\3x-4y-19=0 & 4x+3y-42=0\end{matrix}[/tex]
Menyamakan persamaan
y - y₁ = m(x - x₁)
y - 2 = m(x - 9)
y = mx - 9m + 2
Mengingat persamaan garis singgung lingkaran (x - a)² + (y - b)² = r² yang bergradien m adalah [tex]\displaystyle y-b=m(x-a)\pm r\sqrt{m^2+1}[/tex]
[tex]\displaystyle y-3=m(x-2)\pm 5\sqrt{m^2+1}\\y=mx-2m+3\pm 5\sqrt{m^2+1}\\mx-9m+2=mx-2m+3\pm 5\sqrt{m^2+1}\\-7m-1=\pm 5\sqrt{m^2+1}[/tex]
(-7m - 1)² = 25(m² + 1)
49m² + 14m + 1 - 25m² - 25 = 0
24m² + 14m - 24 = 0
12m² + 7m - 12 = 0
(4m - 3)(3m + 4) = 0
m₁ = ¾ atau m₂ = -4/3
Persamaan nya
[tex]\displaystyle \begin{matrix}y-y_1=m_1(x-x_1) & y-y_1=m_2(x-x_1)\\ y-2=\frac{3}{4}(x-9) & y-2=-\frac{4}{3}(x-9)\\4y-8=3x-27 & 3y-6=-4x+36\\3x-4y-19=0 & 4x+3y-42=0\end{matrix}[/tex]
Metode garis polar
(x₁ - a)(x - a) + (y₁ - b)(y - b) = r²
(9 - 2)(x - 2) + (2 - 3)(y - 3) = 25
7x - 14 - y + 3 - 25 = 0
7x - y - 36 = 0
y = 7x - 36 ← persamaan garis polar
Substitusi ke persamaan lingkaran
(x - 2)² + (y - 3)² = 25
(x - 2)² + (7x - 39)² - 25 = 0
x² - 4x + 4 + 49x² - 546x + 1521 - 25 = 0
50x² - 550x + 1500 = 0
x² - 11x + 30 = 0
(x - 5)(x - 6) = 0
x₂ = 5 atau x₃ = 6
y₂ = 7(5) - 36 = -1 atau y₃ = 7(6) - 36 = 6
Titik singgung nya (5, -1) dan (6, 6)
Persamaan nya
[tex]\displaystyle \begin{matrix}(x_2-a)(x-a)+(y_2-b)(y-b)=r^2 & (x_3-a)(x-a)+(y_3-b)(y-b)=r^2\\ (5-2)(x-2)+(-1-3)(y-3)=25 & (6-2)(x-2)+(6-3)(y-3)=25\\ 3x-6-4y+12-25=0 & 4x-8+3y-9-25=0\\ 3x-4y-19=0 & 4x+3y-42=0\end{matrix}[/tex]
Metode diskriminan
y - y₁ = m(x - x₁)
y - 2 = m(x - 9)
y = mx - 9m + 2
Substitusi ke persamaan lingkaran
(x - 2)² + (y - 3)² = 25
(x - 2)² + (mx - 9m - 1)² - 25 = 0
x² - 4x + 4 + m²x² + 81m² + 1 - 18m²x - 2mx + 18m - 25 = 0
(m² + 1)x² + (-18m² - 2m - 4)x + 81m² + 18m - 20 = 0
D = 0
(-18m² - 2m - 4)² - 4(m² + 1)(81m² + 18m - 20) = 0
324m⁴ + 4m² + 16 + 72m³ + 144m² + 16m - 4(81m⁴ + 18m³ - 20m² + 81m² + 18m - 20) = 0
324m⁴ + 4m² + 16 + 72m³ + 144m² + 16m - 324m⁴ - 72m³ + 80m² - 324m² - 72m + 80 = 0
-96m² - 56m + 96 = 0
12m² + 7m - 12 = 0
(4m - 3)(3m + 4) = 0
m₁ = ¾ atau m₂ = -4/3
Persamaan nya
[tex]\displaystyle \begin{matrix}y-y_1=m_1(x-x_1) & y-y_1=m_2(x-x_1)\\ y-2=\frac{3}{4}(x-9) & y-2=-\frac{4}{3}(x-9)\\4y-8=3x-27 & 3y-6=-4x+36\\3x-4y-19=0 & 4x+3y-42=0\end{matrix}[/tex]