Area $A$ is defined as the area of intersection between $x^2+x-2$ and $3x+1$. If a circle $x^2+y^2=r^2$ fits inside the area $A$, then what is the range $0 < r < \dots$?
At first, since it explicitly says the circle is in the origin, I search for the nearest $x$ or $y$ intercept and found $-1/3$ as the nearest thus its not possible to create a circle with radius more than $1/3$ included in $A$. But the answer is $\frac{1}{10^{1/2}}$, how can I reach this answer?
On
You just have to compute the distance of the origin from the line $y=3x+1$ and from the parabola $y=x^2+x-2$. The first distance is $\frac{1}{\sqrt{10}}$, the second one can be found by identifying the minimum of $$ d^2(x) = x^2+(x^2+x-2)^2 $$ which occurs at a root of a cubic equation. In a simpler way, you may just check that the circle centered at the origin with radius $\frac{1}{\sqrt{10}}$ does not intersect the parabola.
The circle is tangential to either the line or the parabola.
Given that the origin (0,0), the center of the circle, is closer to the line than to the parabola, the upper limit of the circle is just the distance from the origin to the line, which is given by
$$r \le \frac{m}{\sqrt{1+k^2}}=\frac{1}{\sqrt{10}}$$
where $k=3$ is the slope of the line and $m = 1$ is its $y$-intercept.
What you found, -1/3, is the $x$-intercept of the line. The circle can not pass this point, because the line would cut cross the circle and the circle would be partially outside the region $A$.