Radius-4 circle skewered on one line and touching another Find the equation of the circle with radius $4$ units, whose centre lies on the line $4x+13y=32$ and which touches the line $4x+3y+28=0$.
My Approach:
Radius $r=4$ units
Let $P(h,k)$ be the centre of the circle. Then $4h+13k=32$.
Please help me to move further.

The equation of a circle with center in $(a,b)$ and radius $4$ is
$$
\left( {x - a} \right)^{\,2}  + \left( {y - b} \right)^{\,2}  = r^{\,2}  = 16
$$
Now you must have
$$
\left\{ \begin{gathered}
  4a + 13b = 32\quad \text{(center}\,\text{on}\,\text{the}\,\text{line}\,(\text{a))} \hfill \\
  4x + 3y + 28 = 0\quad \text{(point}\,\text{on}\,\text{the}\,\text{line}\;\text{(b))} \hfill \\
  \left( {x - a} \right)^{\,2}  + \left( {y - b} \right)^{\,2}  = 16\quad \text{(point}\,\text{on}\,\text{the}\,\text{circle)} \hfill \\ 
\end{gathered}  \right.
$$
Now, from the first express e.g. $a$ in function of $b$, and from the second e.g. $y$ in function of $x$ and place them in the third. You will get 
$$\left( {x - 8 - \frac{{13}}
{4}b} \right)^{\,2}  + \left( { - \frac{{28}}
{3} - \frac{4}
{3}x - b} \right)^{\,2}  = 16
$$
Expand as a quadratic equation in $x$ and impose to have two coincident solutions, i.e. find $b$ such that the discriminant be $0$.
You will find two values, corresponding to whether the circle is on one side or the other with respect to the crossing point of the two lines.
Use each value of $b$ to determine the corresponding $x$, $y$, $a$.
