Do you need to reproduce this drawing?
First, draw AB with a ruler.
Take your compass, draw an arc with length AB with centre A.
Do the same with centre B.
Where the arcs cut, it is the vertex C.
Then, draw AC and BC with a ruler.
Bisect any of the two sides of the triangle and join the lines of bisection.
Where they meet indicate the centre of the triangle and at the same time the centre of the circle.
Put your compass on the centre of the circle to point D, which is halfway between A and B, (or measure 4.8 units) and draw a circle.
There you are.
I hope that it's what you were looking for.
Well, I was assuming it was an equilateral triangle, because if I think about it, you can have various situations where AC and BC are 17, the circle inscribed in the triangle but various lengths of AB.
Ok, let's get to calculations then.
The let the angle FSE = theta.
Then, angle FCE = 180 - theta
First conclusion: $\displaystyle CS + DS = 17 cos (\frac{\theta}{2})}$ ...1
Elaborating CS + DS: $\displaystyle CS + 4.8 = 17 cos (\frac{180- \theta}{2})$
Elaborating CS: $\displaystyle CS = \frac{4.8}{cos(\frac{\theta}{2})}$
Using what we just found to get to the first conclusion (ie replace the values of CS and DS into 1);
$\displaystyle \frac{4.8}{cos(\frac{\theta}{2})} + 4.8 = 17 cos (\frac{180- \theta}{2})$
$\displaystyle 4.8 + 4.8cos(\frac{\theta}{2}) = 17 cos (\frac{180- \theta}{2})cos(\frac{\theta}{2})$
$\displaystyle 4.8 + 4.8cos(\frac{\theta}{2}) = 17 cos (\frac{180- \theta}{2})cos(\frac{\theta}{2})$
But now, cos(180-theta) = sin(theta)
So,
$\displaystyle 4.8 + 4.8cos(\frac{\theta}{2}) = 17 sin (\frac{\theta}{2})cos(\frac{\theta}{2})$
$\displaystyle 4.8 = 17 sin (\frac{\theta}{2})cos(\frac{\theta}{2})- 4.8cos(\frac{\theta}{2}) $
$\displaystyle 4.8 = 17 cos (\frac{\theta}{2})(sin(\frac{\theta}{2})-4.8) $
Well, I'll use wolframalpha to find that angle.
http://www.wolframalpha.com/input/?i=4.8+%3D+17+cos+(\frac{x}{2})(sin(\frac{x}{2})-4.8)
So, theta = 3.14 rad, which is 174.4 degrees.
So, angle FCE = 5.58 degrees.
From here, you can draw AC first, draw the angle, draw BC, then join AB, and then to find the centre for the circle, find the intersection of the lines DC and AE or any other pair. Open your compass at 4.8 units and draw your circle.
I hope that's it.
The bad news first: I can't show you how to construct the triangle without doing some calculations.
1. The triangle in question is an isosceles triangle with the base b, the area a and the height (2r + x). Compare the attached sketch.
2. The area of the triangle is calculated by:
$\displaystyle a = \frac12 \cdot b \cdot (2r + x)$
Divide the triangle into 3 triangles with the common vertex S. Then the heights of the 3 triangles are the radius r = 4.8. Then the area of the complete triangle is calculated by:
$\displaystyle a = 2 \cdot \frac12 \cdot 17 \cdot r + \frac12 \cdot b \cdot r$
3. The indicated triangle is a right triangle. Use Pythagorean theorem:
$\displaystyle (2r + x)^2+\left(\frac12 b\right)^2=17^2$
4. Solve the system of equations for (a, b, x):
$\displaystyle \left|\begin{array}{rcl}\frac12 b (9.6+x)&=&a \\ 81.6+2.4 b &=& a \\ (9.6+x)^2+\frac14 b^2 &=& 17^2 \end{array}\right.$
5. This system has 4 solutions. Neglecting the solutions with a = 0 or negative side length you'll get 2 plausible solutions:
$\displaystyle (a,b,x)=\left(120, 16, \frac{27}5 \right)$ ........ or
$\displaystyle (a,b,x)=\left(\frac{36}{25} \sqrt{769} + \frac{516}5, \frac{3}{5} \sqrt{769} + 9 , \frac{ \sqrt{769}}2 - \frac{123}{10} \right)$