BCEF is a trapezium. So, the circle should lie inside the trapezium
BCEF is atrapezium. For, AF/BF=9/12=3/4. AE/CE=6/8=3/4. Therefore, AF/BF=AE/CE. Therefore, EF is parallel to AB. Hence, BCEF is a trapezium.
For trapezium, the diameter of the incircle is height of the trapezium. Therefore, radius=height/2 of the trapezium. If BC is known, it is easy to find the radius.
InCircle means circle inside a polygon touching all sides of the polygon or each side of the polygon is a tangent to the circle.
We have to split the problem into three sub cases. Case1: Right-angled triangle, with hypotenues=AB, then radius of the circle = 8, if we change the hypotenus as BC, the result will change, similarly, if we change the hypotenuse as AC, the result will change to 12.
Case2: Isosceles triangle
If we assume, AB = BC, then radius of the circle=80*sqrt(6)/3. If we assume AC=BC, the result will change.
Case3: Scalene triangle
In this case, the answer will be in terms of angle, so exact value is not known to me.