If the line joining the points $(0, 3)$ and $(5, -2)$ is the tangent to the curve ;$\displaystyle y=\frac{c}{x+1}$ then the value of $c$ is

Slop of line joining the points $(0,3)\quad and\quad (5,2)$ is given by

$\dfrac { 3+2 }{ 0-5 } =-1$

This is equal to the slop of tangent on the curve and that is given by

$\dfrac { dy }{ dx } =\dfrac { -c }{ { \left( x+1 \right) }^{ 2 } } \\ \dfrac { dy }{ dx } =-1\\ \Rightarrow c={ \left( x+1 \right) }^{ 2 }...................(1)$\

Equation of line joining the points $(0,3)\quad and\quad (5,2)$ is given by

$(y-3)=-1(x-0)$

Solving equation of tangent and the curve for point of intersection

$\dfrac { c }{ x+1 } +x=3$.............(2)

Solving (1) and (2)

$x=1$

Putting this in (2), we get c=4