The long hand turns 6 degrees a minute and the short one half a degree, so a 90 degree gap closes at 5.5 degrees a minute and the hands coincide 180/11 minutes past three, at 3:16:21.8181. At 3:15 the long hand has reached the 3 and the short hand is 7.5 degrees ahead of it, which is the whole content of the wrong answer. Consecutive coincidences are 720/11 minutes apart, so there are eleven per twelve hours and twenty-two per day, and the common phrasing "eleven times a day" is wrong by a factor of two.
Running straight out from the centre loses, because a radius costs you 1 while half the fence costs the dog pi over 4. Inside a quarter of the radius your angular speed beats his, so you can orbit until he is diametrically opposite and then sprint three quarters of a radius against his pi over 4, and the whole escape reduces to 3 being less than pi with a margin of 0.0354 R. That two-phase plan works only up to a speed ratio of pi + 1, while the best known strategy for the problem reaches 4.60334.
Differentiating y = L tan(wt) gives a spot speed of w R squared over L, so the footprint accelerates with the square of its distance from the lamp: a tenth of pi directly opposite, and exactly pi miles per second nine miles along. The 9 is the along-shore leg, which makes 90 the squared hypotenuse rather than the square of nine, and that misreading is the usual failure. Nothing physical moves at that speed, and a straight coast running 2310 miles would carry a nominally faster-than-light spot carrying no information at all.