Values of constant $k$ for which line $2x+ky =-k$ passes within $1$ unit of origin.

Question

The problem is taken from the chap. 1.1 of the book titled : Calculus Problems for the new century, by Robert Fraga.

The equation $2x+ky =-k$ is the equation of a line. Determine the values of the constant $k$, if any, for which the line passes within one unit of the origin.

The book has solutions to the problems, but am unable to understand it as given below:

$2x+ky =-k\implies y = -1-\frac2kx\implies m= -\frac2k.$
Slope of perpendicular to the given line is $m_1= \frac k2.$
The intersection of this line with the original one when $ -1-\frac2kx=\frac k2x\implies x = -\frac{2k}{k^2+4}$. The $y$-coordinate of this point is $-\frac{k^2}{k^2+4}$.
This point lies within one unit of origin if $x^2+y^2 = \frac{(4k^2+k^4)}{(k^2+4)^2}\le 1$
which simplifies to $4k^2\ge -16$, so the line always passes within $1$ unit of the origin!
This result can be easily be seen from the original equation since the line passes through $(0,-1)$ which is exactly $1$ unit from the origin.

I am unable to find out the below from the answer given:

1. Why perpendicular line is taken?

2. Also, why the $y$-coordinate of the intersection point is taken by multiplying $x = -\frac{2k}{k^2+4}$ with $m_1$, not $m$.

3. $\frac{(4k^2+k^4)}{(k^2+4)^2} = \frac{k^2(4+k^2)}{(k^2+4)^2}$
   Take common factor $(4+k^2)$, & eliminate from numerator & denominator to get: $\frac{k^2}{(k^2+4)}$
   So, get $\frac{k^2}{(k^2+4)}\le 1$, but this is not the same as $4k^2\ge -16$.
   Although, the correct inequality gives the same result as:
   $k^2\le k^2+4\implies 0 \le 4$, which is always true.

Answer 1

1. The solution computes the distance of the line to the origin. This distance is realized by the line that is perpendicular to the original line and passes through the origin.
2. At $x=-\frac{2k}{k^2+4}$ the two lines intersect. To compute the $y$ coordinate of the intersection one can plug $x$ into either one of the two line equations. $y=m_1 x$ is a slightly easier computation.
3. The author brought the $(k^2+4)^2$ to the other side of the equation and then multiplied it out. Both your and his way result in the same, ie all values of $k$ work.