Solution: Connect the AC power supply.

Then < ACB = 90.

Do CE⊥AB at point e

Then BC? =BE*BA

∴BE=x？ /2R

∴CD=2R-2BE=2R-x？ /R

∴y=2AD+AB+CD=2R+2x+2R-x？ /R

∴y=-( 1/R)x？ +2x+4R

(2)

If there is no graph, there should be two situations.

When AD is in △ABC, this triangle is an acute triangle.

Then the radius r of the circle that can completely cover the triangle ABC is the radius of the circumscribed circle.

13 *15 = 2r *12 (the product of any two sides of a triangle is equal to the height of the third side multiplied by the diameter of the circumscribed circle).

R=65/8

When AD is outside the triangle, the triangle is obtuse.

Then the radius r of the circle that can completely cover the triangle ABC is half of the longest side, that is, 15/2.