Let $\mathrm{y}=\mathrm{x}+2, 4\mathrm{y}=3\mathrm{x}+6 \mathrm{and} 3\mathrm{y}=4\mathrm{x}+1$be three tangent lines to the circle ${(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2$ . Then$\mathrm{h}+\mathrm{k}$ is equal to:

$\left(\mathrm{h},\mathrm{k}\right) \mathrm{is} \mathrm{the} \mathrm{incentre} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{formed} \mathrm{by} \mathrm{the} \mathrm{given} \mathrm{lines}$

$$\begin{gathered} \text{ Bisector of lines }{\mathrm{L}}_2\mathrm{\&}{\mathrm{L}}_2 \\ \frac{4\mathrm{x}-3\mathrm{y}+1}{5}=\pm \left(\frac{3\mathrm{x}-4\mathrm{y}+6}{5}\right) \end{gathered}$$

$\text{ (+) }4x-3y+1=3x-4y+6$

$x+y=5$

In centre lines on $x+y=5$ therefore h+k=5

$\mathrm{AB}=\sqrt{49+49}=7\sqrt{2}$
$\begin{array}{l}\mathrm{BC}=\sqrt{9+16}=5\\ \mathrm{CA}=\sqrt{16+9}=5\end{array}$