Persamaan lingkaran yang berpusat (-2,1) Dan menyinggung garis 5x+8y-4=0 adalah

tentukan jarak titik pusat ke garis.

x - 3y - 4 = 0
3y = x - 4
y = (1/3)x - 4/3

m1 = 1/3
m2 = -3

y - b = m(x - a)
y -(-4) = -3(x - 2)
y + 4 = -3x + 6
y = -3x + 2

y = y
(1/3)x - 4/3 = -3x + 2
x - 4 = -9x + 6
10x = 10
x = 1

y = -3x + 2
y = -3(1) + 2
y = -1

r^2 = ((1 - 2)^2 + (-1 -(-4))^2)
= (-1)^2 + (3)^2
= 1 + 9
= 10

(x - a)^2 + (y - b)^2 = r^2
(x - 2)^2 + (y - (-4))^2 = 10
(x - 2)^2 + (y + 4)^2 = 10
x^2 - 4x + 4 + y^2 + 8y + 16 - 10 = 0
x^2 + y^2 - 4x + 8y + 10 = 0