Solving for equation of a circle when tangents to it are given. Given: $\mathrm L(x) = x - 2$, $\mathrm M(y) = y - 5$ and $\mathrm N(x, y)= 3x - 4y - 10$.
To find: All the circles that are tangent to these three lines.

Outline of the method :
If we parameterised any line $\mathrm Z$ in terms of $x(t) = pt + q$ and $y(t) = rt + s$, then for a general circle $\mathrm C(x, y) = (x - \alpha)^2 + (y - \beta)^2 - \gamma^2$ we can substitute the the parameters of the line and get
$$\mathrm C(x(t), y(t))  = a t^2 + bt + c  = 0,$$
where $a,b,c \in \Bbb R$ and they depend on $p$,$q$.
Now if this quadratic has $2$ roots then the circle intersects the line at $2$ points and same for $1$ and $0$ roots.
If we are given that $\mathrm Z$ is tangent then we can using completing the square and get $(t - d(p, q) ) ^2  + e(p, q) = 0$.
So our condition for the line tangent to the circle would be $e(p, q) = 0$.

Parameterising the three equations, 
$\mathbf L^\prime (l) = (2,0) + l (0,1)$, $\mathbf M^\prime(m) = (0, 5) + m(-1, 0)$ and $\mathbf N^\prime(n) = (2 , -1) + n(4,3)$.
Substituting these in the general equation of a circle,  $\mathrm C(x, y) = (x - h)^2 + (y - k)^2 - r^2$.
$$\left\{ \begin{align} \mathrm{C(L_x, L_y)} &= (2-h)^2 +(l-k)^2 -r^2 \\ \mathrm{C(M_x, M_y)} &= (m+h)^2 +(5-k)^2 -r^2 \\ \mathrm{C(N_x, N_y)} &= (2+4n-h)^2 +(3n-1-k)^2 -r^2 \end{align} \right.$$
As per the above method, 
From first equation we get $r^2 = (2 - h) \iff r = |2 - h|$ and from second equation, $r = |5 - k|$.
Completing the square on the third equation we get,
$$\mathrm{C(N_x, N_y)} = \left(5n + \dfrac{5 - 4h - 3k}{5}\right) + 5 - 4h + 2k - r^2 + h^2 + k^2 - \left[1 - \dfrac{4h - 3k}{5}\right]^2.$$ 
Therefore,
$$ 2 + 2k + h^2 + k^2 = r^2 + 4h +  \left[1 - \dfrac{4h - 3k}{5}\right]^2.$$
So here we got three equations:
\begin{align} 2 +2k +h^2 +k^2 &= r^2 +4h +\left[1 - \dfrac{4h-3k}{5}\right]^2 \\ r &= |5-k| \\ r &= |2-h|. \end{align}
I don't mind solving the third equation after substituting for $r$ from first in it, it is just a quadratic. What I don't understand is whether I should take $r = + (5 - k)$ or $r = - (5 - k)$ and $r = +(2 - h)$ or $r = - (2 - h)$?
Which two out of those should I take? And why?  
 A: converting the equation in the Hessian Normalform we have
$$\frac{|3x-4y-10|}{5}=R$$ where $$(x,y)$$ denotes the middle Point of our searched circle; and it must be $$x=2+R,y=5-R$$
A: Consider $\triangle ABC$,
$A=(2,5),B=(2,-1),C=(10,5)$
and its incircle and three excircles,

\begin{align}
a&=10,\quad
b=8,\quad
c=6,\quad
\\
r&=2,\quad
r_a=12,\quad
r_b=6,\quad
r_c=4
,\\
O_i&=(4,3),\quad
O_a=(14,-7),\quad
O_b=(8,11),\quad
O_c=(-2,1)
.
\end{align}  
A: You have by distance from a point to a line formula,
$$|h-2|=|k-5|=\frac{|3h-4k-10|}{5}=r.$$
Consequently, $h=2\pm r$, and $k=5\pm r$. So we have four cases


*

*$h=2+r$, $k=5+r$ implies $|3(2+r)-4(5+r)-10|=|r+24|=5r \implies r=6$. So the equation of the circle is $$\color{red}{(x-8)^2+(y-11)^2=36}.$$

*$h=2+r$, $k=5-r$ implies $|3(2+r)-4(5-r)-10|=|7r-24|=5r \implies r=2,12$. So the equation of the circles are $$\color{red}{(x-4)^2+(y-3)^2=4},$$ and $$\color{red}{(x-14)^2+(y+7)^2=144}.$$

*$h=2-r$, $k=5+r$ implies $|3(2-r)-4(5+r)-10|=|7r+24|=5r$, which does not have any solution such that $r >0$.

*$h=2-r$, $k=5-r$ implies $|3(2-r)-4(5-r)-10|=|r-24|=5r \implies r=4$. So the equation of the circle is $$\color{red}{(x+2)^2+(y-1)^2=16}.$$

