Let $f(x, y)=ax+by+c$. Find the largest value of $r$ such that $f(x, y)>0$ for all pairs $(x, y)$ satisfying $x^2 + y^2 < r^2$? 
Let $f(x, y) = ax + by + c$, where $\{a, b, c\} \subset\mathbb R$ and $c > 0$. Find the largest value of $r$ such that $f(x, y) > 0$ for all pairs $(x, y)$ satisfying $x^2 + y^2 < r^2$.

I was trying this question many times, but I could not find it.  I don't know where i have to start.
I was taking taking the radius $r= 1$, $x= \frac{1}{2}$ and $y =\frac{1}{2}$, possisbly it is meaningless to take this value because it never less than $1$ as $\frac{1}{2} + \frac{1}{2}= 1$ so it is contradict.
If anybody help me I would be very thankful to him.

By C-S
$$r^2>x^2+y^2=\frac{(a^2+b^2)(x^2+y^2)}{a^2+b^2}\geq\frac{(ax+by)^2}{a^2+b^2}.$$
Thus, $$-r\sqrt{a^2+b^2}<ax+by<r\sqrt{a^2+b^2}$$ or
$$c-r\sqrt{a^2+b^2}<ax+by+c<c+r\sqrt{a^2+b^2}.$$
Thus, we need $$c-r\sqrt{a^2+b^2}\geq0$$ or
$$r\leq\frac{c}{\sqrt{a^2+b^2}}.$$
