Say I'm given two lines $L_1:a_1x+b_1y+c_1=0$ and $L_2:a_2x+b_2y+c_2=0$ tangent to a circle of radius $r$. If I proceed to find the coordinates of the centre $(h,k)$, I could use the perpendicular-distance-of-a-point-from-a-line formula i.e. $$d=\frac{|a(p)+b(q)+c|}{\sqrt{a^2+b^2}}$$ for a point$(p,q)$ from a line $ax+by+c=0$, $d$ being the distance between them.
I could say the following;
$r=\frac{|a_1(h)+b_1(k)+c_1|}{\sqrt{a_1^2+b_1^2}}=\frac{|a_2(h)+b_2(k)+c_2|}{\sqrt{a_2^2+b_2^2}}$
from general properties of circles and tangents.
My question is, given a condition, (say, 'The centre lies in first quadrant') which two operations out of $±$ do I use?
More precisely, out of the four pairs of equations that the following represents, how do I save from solving each of them and seeing for both positive values for $h$ and $k$? i.e. how do I know which pair to solve?
$\frac{a_1(h)+b_1(k)+c_1}{\sqrt{a_1^2+b_1^2}}=±r=\frac{a_2(h)+b_2(k)+c_2}{\sqrt{a_2^2+b_2^2}}$
Or
$±r=\frac{a_1(h)+b_1(k)+c_1}{\sqrt{a_1^2+b_1^2}}$ And $±r=\frac{|a_2(h)+b_2(k)+c_2|}{\sqrt{a_2^2+b_2^2}}$
Note first that there are four different circles of a given fixed radius $r$ tangent to the two lines.
Given a point $(h,k)$ and a line $ax + by + c = 0$, if we assume that $b>0$ (which we can do without loss of generality), then $ah+bk+c$ is positive or negative depending on whether $(h,k)$ is above or below the line, respectively.
So, instead of specifying which quadrant the center is in (which wouldn't determine the signs anyway), what you really want is to specify where the center is relative to the two lines: above both, above the first and below the second, below the first and above the second, or below both. Each of these four choices corresponds to one of the four circles. Then you can set \begin{equation} r = \pm \frac{a_i h+b_i k+c_i}{\sqrt{a_i^2+b_i^2}} \end{equation} for $i=1,2$ accordingly, and solve for $(h,k)$ using the pair of linear equations.