Find the coefficient "k" such that the line  kx + y + 1 = 0 
 
     is tangent to the circle of the radius    centered at the point (2,1)

The standard form equation of the circle is

     +  = 3.           (1)


The equation of the line is

    y = -kx - 1.                      (2)



Substitute the equation (2) into equation (1), replacing y in the equation (1).

Doing this way, you will get then a single equation for only one unknown x
    
     +  = 3     (3)

or

     +  = 3.          (4)



The idea of the solution to the problem is to reduce equation (4) to standard form quadratic equation
and then to find the value of "k" from the condition that this equation has ONLY ONE real solution
(which is equivalent to the fact that the line(2) and the circle (1) have ONLY ONE common point).


So, we simplify equation (4) step by step

    x^2 - 4x + 4 + k^2x^2 + 4kx + 4 = 3

    x^2(1+k^2) - 4x(1-k) + 5 = 0.       (5)


Now we will find "k" from the condition that the discriminant of equation (5) is equal to zero.


The discriminant  d  is

    d = b^2 - 4ac =  -  =  - .


The condition d = 0  is

        = 

    4(1-k)^2       = 5(1+k^2)

    4 - 8k + 4k^2  = 5 + 5k^2

    k^2 + 8k + 1 = 0

     =  =  = .


So, there are 2 (two, TWO) values for k:   =   and   = .


ANSWER.  There are two solutions for "k" :   =   and   = .