Jawab:
Persamaan garis x = 3, dan 3x-4y+11 = 0
Titik singgung      (3, 1)  dan   (-¹/₅, ¹³/₅)

Penjelasan dengan langkah-langkah:
Dik persamaan
x² + y² - 2x - 2y - 2 = 0
Cek posisi titik (3, 5) thd persamaan
x = 3, y = 5, maka
3² + 5² - 2(3) - 2(5) - 2 = 0
9 + 25 - 6 - 10 - 2 = 0
16  >  0
Nilai lebih dari 0 maka titik berada diluar
lingkaran.

Cari titik tengah (P, Q) dan jari-jarinya (r)
x² + y² - 2x - 2y - 2 = 0
x² + y² + ax + by + c = 0
a = -2, b = -2, c = -2

Titik tengah             jari-jari (r)
(P, Q) = (-a/2, -b/2),  4r² = (a²+b²-4c)

(P, Q) = (-(-2)/2, -(-2)/2),  4r² = ((-2)²+(-2)²-4(-2))
(P, Q) = (2/2, 2/2) ,  4r² = (4+4+8)
(P, Q) = (1, 1) ,  4r² = 16
(P, Q) = (1, 1) ,  r² = 16/4
(P, Q) = (1, 1) ,  r² = 4
(P, Q) = (1, 1) ,  r = √4
(P, Q) = (1, 1) ,  r = 2

(x-P)(x₁-P) + (y-Q)(y₁-Q) = r²
(x-1)(x₁-1) + (y-1)(y₁-1) = 4
x₁ = 3, y₁ = 5, maka
(x-1)(3-1) + (y-1)(5-1) = 4
2(x-1) + 4(y-1) = 4
2x - 2 + 4y - 4 = 4
2x + 4y = 4 + 4 + 2
2x + 4y = 10
2(x+2y) = 2(5)
x + 2y = 5
x = 5 - 2y

Lalu cari titik potong lingkaran
(x-P)² + (y-Q)² = r²
(x-1)² + (y-1)² = 4
((5 - 2y)-1)² + (y-1)² = 4
(4-2y)² + (y-1)² = 4
4² - 2(4)(2y) + (2y)² + y² - 2y + 1 = 4
16 - 16y + 4y² + y² - 2y + 1 = 4
5y² - 18y + 17-4 = 0
5y² - 18y + 13 = 0
5y² - 5y - 13y + 13 = 0
5y(y-1) - 13(y-1) = 0
(5y-13)(y-1) = 0
y = ¹³/₅  dan y = 1

Untuk y = 1
x = 5 - 2(1)
x = 5 - 2
x = 3
Titik singgung (x, y) = (3, 1)

Garis yang melalui (3, 5)
adalah x = 3

Cari persamaan satu lagi
Cari x jika y = ¹³/₅
x = 5 - 2(¹³/₅)
x = 5 - ²⁶/₅
x = ²⁵/₅ - ²⁶/₅
x = -¹/₅
Titik singgung (x, y) = (-¹/₅, ¹³/₅)

Cari pers garis
(x-P)(x₁-P) + (y-Q)(y₁-Q) = r²
(x-1)(-¹/₅-1) + (y-1)(¹³/₅-1) = 4
(x-1)(-¹/₅-⁵/₅) + (y-1)(¹³/₅-⁵/₅) = 4
(x-1)(-⁶/₅) + (y-1)(⁸/₅) = 4
5(x-1)(-⁶/₅) + 5(y-1)(⁸/₅) = 4(5)
-6(x-1) + 8(y-1) = 20
-6x + 6 + 8y - 8 = 20
8y = 6x + 20 - 6 + 8
8y = 6x + 22
(8/2)y = (6/2)x + (22/2)
4y = 3x + 11
atau 3x-4y+11 = 0

(xcvi)