Persamaan garis singgung pasa lingkaran x²+y²-4x+6y-12=0 melalui titik (6,-6)

x² + y² + Ax + By + C = 0
x² + y² - 4x + 6y - 12 = 0
Titik (6,-6)

A= -4
B = 6
C = -12
x1 = 6
y1 = -6

x1x + y1y + A ((x1 + x)/2) +B ((y1 + y)/2) + C = 0
6x + (-6)y + (-4) ((6+x)/2) + 6 ((-6 + y)/2) + (-12) = 0
6x - 6y -12 - 2x - 18 + 3y - 12 = 0
4x - 3y - 42 = 0

Lingkaran.
Kelompok peminatan kelas XI kurikulum 2013 revisi 2016.

x² + y² - 4x + 6y - 12 = 0
x² - 4x + 4 + y² + 6y + 9 = 12 + 4 + 9
(x - 2)² + (y + 3)² = 25

Persamaan garis singgungnya
(x - a)(x₁ - a) + (y - b)(y₁ - b) = r²
(x - 2)(6 - 2) + (y + 3)(-6 + 3) = 25
4x - 8 - 3y - 9 - 25 = 0
4x - 3y - 42 = 0