OK. The picture is an isosceles triangle PFT with PT = FT = 12, and PF = 5 1/2. When you draw the line equidistant from F and T, it’s perpendicular to segment FT and through its midpoint, 6 cm from either end. You are correct about it hitting at two points. What they mean is that if the circle had a smaller radius, it would only hit the line from part a once (would be tangent to the line. So they want you to find what distance (smaller than 6) that would be. I don’t know what level math this question is from, but one way to solve it is to draw the original triangle and part a line, but draw the circle centered at P so it’s tangent to the line. Draw the radius to the point of tangency. Call this length x (it’s what they are asking for). Then draw a segment from P that is perpendicular to segment FT, and call the point where it hits FT point G. This gives you a rectangle and a right triangle PFG. You can find FG and subtract from 6; that would be x since it’s on the opposite side of the rectangle from x. Now draw an altitude of the original triangle to point T, splitting the 5 1/2 into two 2 3/4 pieces. These right triangles with a 2.75 leg and 12 hypotenuse, are similar to triangle PFG.

So 2.75/12 = FG / 5.5 .

12 FG = 5.5(2.75) or 15.125 by cross multiplying; divide that by 12 to get FG (some long decimal) then subtract it from 6 to get x.

Wow that WAS tough! Good luck trying to follow all that!