How do I find the equation of a circle given the equation of $3$ tangents I would love some help on finding the equation of the circles tangent to $d_1, d_2$ and $d_3$, given 
$$\begin{cases}d_1: y=4x-10
\\
d_2:  y=9/4x-15/4
\\
d_3: y=3x-15
\end{cases} $$
My approach: I know that $d_2$ and $d_3$ are parallel. The circles, in order to be tangent to the $3$ lines must pass by $M$( intersection between $d_1$ and $d_2$) and $N$ (intersection between $d_1$ and $d_3$).
I found the coordinates of these points. Then, I found the radius, as it is the distance between $ d_2$ and $d_3$. But in the end, it appears that I am wrong, but I dont understand why...
Many thanks in advance

The distance to a straight line of equation $$ax+by+c=0$$ is obtained as
$$\frac{|ax+by+c|}{\sqrt{a^2+b^2}}.$$
The center of a circle tangent to the three lines is equidistant to them and you need to solve
$$\dfrac{|ax+by+c|}{\sqrt{a^2+b^2}}=
\dfrac{|a'x+b'y+c'|}{\sqrt{a'^2+b'^2}}=
\dfrac{|a''x+b''y+c''|}{\sqrt{a''^2+b''^2}}.$$
Removing the absolute values, you have actually four systems of two equations in two unknows, giving four distinct solutions (they are the inscribed circle and three escribed ones).
