For what value(s) of k is the line x+2y+k=0 tangent to the circle x^2+y^2-2x+4y+1=0

x+2y+k=0 :eqn 1
x^2+y^2-2x+4y+1=0: eqn 2
x = -2y -k : eqn 3

to find where the circle and line touch, sub eqn 3 into eqn 2
(-2y-k)^2 + y^2 -2(-2y-k) +4y +1 = 0
4y^2 + 4k y +k^2 + y^2 +4y +2k +4y +1 = 0
5y^2 +(4k +8) y + k^2 + 2k + 1 = 0
y = -(4k +8)/10 +/- sqrt( (4k+8)^2 -4(5)(k^2 + 2k + 1)/10
since the tangent touches the circle at only one point the discriminant must be zero

(4k+8)^2 -4(5)(k^2 + 2k + 1) = 0
16 k^2 + 64 k + 64 -20 k^2 -40 k -20 = 0
4k^2 -24 k -44 = 0
k^2 - 6 k - 11 = 0
k^2 - 6 k - 11 = 0
a = 1
b = -6
c = -11
-b/(2a) =+3
b^2 - 4 a c = 80
sqr((b^2-4ac))/(2a) =sqrt(80)/2 = 2 sqrt(5)
k = -b/(2a) +/- sqr( b^2 -4ac) / (2a)
k = 3 +/- 2 sqrt(5)




D' + 5E' + F' = -26 : eqn 1
-2D' + 3E' + F' = -13: eqn 2
2D' - E' + F' = -5: eqn 3
sub eqn 2 from eqn 1 to eliminate F'
3 D' + 2 E' =-13: eqn 4

sub eqn 2 from eqn 3
4 D' - 4 E' = 8
D' - E' = 2 : eqn 5
add 2 * eqn 5 to eqn 4
5 D' = -9
D' = - 9/5

using eqn 5
-9/5 - E' = 2
E' = -9/5 -2 = -19/5

using eqn3
2(-9/5) - (-19/5) + F' = -5
F' = -25/5 + 18/5 -19/5 = -26/5