At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (9.00i + 8.00j) m/s^2. It moves at constant speed. At time t2 = 3.00s (3/4 of a revolution later), it's acceleration is (8.00i  9.00j) m/s^2. Find the radius of the path taken by the particle.
Constant speed implies no tangential acceleration. The acceleration is centripetal. So you know it has an angular velocity of displacement/time (solve that, you have the numbers above.)
magnitude acceleration= w^2 * r solve for r.
1 answer

r = (9.00i + 8.00j) / (w^2)
w = (t2  t1) / (3/4)
r = (9.00i + 8.00j) / (((3.00  2.00) / (3/4))^2)
r = (9.00i + 8.00j) / (1.00/9.00)
r = (9.00i + 8.00j) * 9.00
r = 81.00i + 72.00j m