#### Answer

$a=-3, b=2$ and $c=1$.

#### Work Step by Step

$y=f(x)=x^2+ax+b$ and $y=g(x)=cx-x^2$
1) Since $g(x)$ has a tangent line at $(1,0)$, the curve $g(x)$ passes through the point $(1,0)$: $$c\times1-1^2=0$$ $$c-1=0$$ $$c=1$$
Therefore, the equation of $g(x)$ is $g(x)=x-x^2$
2) Find $f'(x)$ and $g'(x)$: $$f'(x)=2x+a$$ $$g'(x)=1-2x$$
- At $(1,0)$, the slope of the tangent line to $f(x)$ and $g(x)$ is $$f'(1)=2\times1+a=2+a$$ $$g'(1)=1-2\times1=1-2=-1$$
Since $f(x)$ and $g(x)$ have the same tangent line at $(1,0)$, these two slopes must be equal.
$$2+a=-1$$ $$a=-3$$
3) Finally, since $f(x)$ has a tangent line at $(1,0)$, $f(x)$ must pass through the point $(1,0)$: $$1^2+a\times1+b=0$$ $$1+a+b=0$$
Substitute $a=-3$ here: $$1-3+b=0$$ $$-2+b=0$$ $$b=2$$