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

line x+2y + k =0------------------------------(I)
circle x^2 + y^2 - 2x + 4y +1 =0 -------------(II)
centre of circle is ( 1 ,- 2)& radius = sqrt( 1^2+2^2-1) = 2
[ if x^2+y^2+2gx+2fy+c=0 is circle then it's centre is(-g,-f) & radius = sqrt(g^2+f^2-c) ]
now if (I) is tangent to circle(II) then lengthof perpendicular from centre (1,-2)
to the line (I) will be equal to the radius of circle 2.
so 2 = (1 -4+ k) /sqrt(1^2+2^2)
or k-3 = 2sqrt5
or k= 3 +_ 2sqrt5 ANSWER
-----------------------------OR-------…
if line lx+ my+ n=0 is tangent to circle x^2+y^2+2gx+2fy+c=0
then (-gl-fm+n/sqrt(l^2+m^2) = sqrt( g^2+f^2-c)
OR ( gl+fm-n)^2 = ( l^2+m^2) (g^2+f^2-c)------------------(I)
now in this question g=-1,f=2,c=1, l=1,m= 2& n= k
so putting these values in (I)
(- 1+ 4-k)^2 = (1+4) (1+4-1)
( 3- k)^2 = 20
or 3- k = sqrt20= +_2 sqrt5
so k = 3 +_2sqrt5--------------------------------… 2nd if there are three such equation then it is solved by making determinants
as D/| | = E/| | = F/| | = 1/| |
when you make deteminant then the values are as
D/ -36= E = -76 = F/-104 = 1/20
so D = -36/20 = -9/5
E = - 76/20 = -19/5
F = -104/20 = - 26/5
if you don't know how to make determinant then i will give you an example
for D take coefficients of E in one column,coefficients of F in second column
& constants in third column
that will be
5 1 -26
3 1 -13
-1 1 -5 and so on [ i am unable to write in my laptope othrewise i will give the in details]
-1