Answer
$c=4$
Work Step by Step
Step 1. Given $y=\frac{c}{x+1}$, we have $y'=-\frac{c}{(x+1)^2}$, which gives the slope of tangent lines to the curve.
Step 2. The equation for the line passing the given two points can be written as $\frac{y+2}{3+2}=\frac{x-5}{0-5}$ or $y=-x+3$, which has a slope of $m=-1$
Step 3. For the line to be a tangent to the curve, we have $y'=m$ or $-\frac{c}{(x+1)^2}=-1$ or $c=(x+1)^2$, where $x$ is the value from the intersect point.
Step 4. To find the intersect point, plug the line equation to the curve equation; we have $\frac{c}{x+1}=-x+3$ or $c=-(x-3)(x+1)$
Step 5. Comparing the results from steps 3 and 5, we have $(x+1)^2=-(x-3)(x+1)$. As $x\ne1$, we have $x+1=-x+3$ so that $x=1$ and $c=4$
