X=2+ 1=3 (optimal arc)

2*2-3= 1 (bad arc)

Center o, chord left end point a, chord right end point b and lower arc midpoint c.

Connect OA and OB, then OA=2.

Connecting OC and AB to P is ∠AOC=∠BOC, and it is easy to get △OPA≠△OPB. So the vertical division of OC AB and P is the midpoint of AB.

Get OP= 1。

All I want is a PC, the distance is 1 cm.

2.

O passing through the center of the circle is perpendicular to the chord ab, and the vertical foot is C, then oc is the distance from the center of the circle to AB.

Connecting ao, bo, triangle aob is isosceles triangle, aoc is right triangle.

Derived from Pythagorean theorem

oc^2=5^2-3^2

So oc=4

3. If the chord length of the bow is 24cm and the height of the bow is 8cm, then the diameter of the circle where the bow is located is-26.