Given the circle C: x ^ 2 + y ^ 2-8y + 12 = 0, the line L; KX + y + 2K = 0, when the value of K is, the line L is tangent to C

Given the circle C: x ^ 2 + y ^ 2-8y + 12 = 0, the line L; KX + y + 2K = 0, when the value of K is, the line L is tangent to C

The formula of the circle equation is x ^ 2 + (y-4) ^ 2 = 4, so the center coordinate of the circle is (0,4), and the radius r = 2. Because the line is tangent to the circle, the distance from the center of the circle to the line = radius, that is | 2K + 4 | / √ (k ^ 2 + 1) = 2. Remove the denominator and square both sides to get 4K ^ 2 + 16K + 16 = 4K ^ 2 + 4, and the solution is k = - 3 / 4