#### Answer

The answer we need to find is $c=4$.

#### Work Step by Step

$$y=\frac{c}{x+1}$$
1) Find the derivative of $y$: $$y'=\frac{c'(x+1)-c(x+1)'}{(x+1)^2}=\frac{0(x+1)-c(1)}{(x+1)^2}$$ $$y'=-\frac{c}{(x+1)^2}$$
2) We need to find the equation of the mentioned line, whose form is following $$y=ax+b$$
- Since the line goes through $(0,3)$: $$a\times0+b=3$$ $$b=3$$
- Since the line goes through $(5,-2)$: $$5a+b=-2$$
Apply $b=3$: $$5a+3=-2$$ $$5a=-5$$ $$a=-1$$
So the mentioned line is $y=-x+3$
3) A line is tangent to a curve when it intersects the curve at one and only one point, or in the words, the equation $$\frac{c}{x+1}=-x+3$$ has only one solution.
$$\frac{c}{x+1}=-x+3$$ $$c=(-x+3)(x+1)=-x^2-x+3x+3$$ $$c=-x^2+2x+3$$ $$x^2-2x+c-3=0$$
Recall that for a quadratic equation $y=Ax^2+Bx+C$, we have $$\Delta=B^2-4AC$$
The equation has only one solution when $\Delta=0$.
Here, in this case, $A=1$, $B=-2$ and $C=c-3$ $$\Delta=(-2)^2-4\times1\times(c-3)$$ $$=4-4(c-3)=4-4c+12=16-4c$$
Then, for $\Delta=0$, $$16-4c=0$$ $$c=4$$
So the answer we need to find is $c=4$.