This is how I approached the problem:

Bisect the isosceles triangle into two right triangles, and put one of the right triangles in the $xy$-plane with the right angle at the origin, as in the diagram:

The vertical leg has length $b$ and the horizontal leg has length $b$.

The hypotenuse of the triangle lies along the line $y(x)$. By the two-intercept form of a line, we know:

$ \displaystyle \frac{x}{a}+\frac{y}{b}=1$

By Pythagoras, and the fact that the hypotenuse is 1 unit in length, we know:

$ \displaystyle a^2+b^2=1$

This will allow you to write the line in terms of one parameter.

Next, use the fact that the perpendicular distance from the point $(r,0)$ to this line is simply $r$. This will allow you to write $r$ as a function of one parameter. If you express the line in slope-intercept form, you may use the formula for the perpendicular distance between a point and a line:

$ \displaystyle d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}$

Once you have $r$ as a function of one parameter, differentiate $r$ with respect to this parameter, and equate to zero to find the critical value within the domain. Then evaluate $r$ at this critical value to find the maximum radius, and plug this into the formula for the area of the circle having this radius.

Please feel free to post your work if you get stuck or need further guidance.