Let y = x + 2, 4y = 3x + 6 and 3y = 4x + 1 be three tangent lines...

Lines $y = x + 2$, $4y = 3x + 6$ and $3y = 4x + 1$ are tangents to the circle $( x - h{)}^{2} + ( y - k{)}^{2} = {r}^{2}$.

Centre of the circle is \left(h, k\right)$.</p><p>Equation of bisector of lines$4y = 3x + 6, 3y = 4x + 1$is:</p><p>$\dfrac{4x - 3y + 1}{5} = \pm \left(\dfrac{3x - 4y + 6}{5}\right)$</p><p>$⇒ 4x - 3y + 1 = \pm \left(3x - 4y + 6\right)$</p><p>Taking positive sign, we get</p><p>$4x - 3y + 1 = 3x - 4y + 6$</p><p>$⇒ x + y = 5$</p><p>Since, centre \left(h, k\right)$ lies on the bisector, therefore

$h + k = 5$