The fixed point a (3,0) is a point outside the circle x ^ 2 + y ^ 2 = 1, P is a moving point on the circle, the bisector of ∠ POA intersects Pa at Q, and the trajectory of Q point is obtained

The fixed point a (3,0) is a point outside the circle x ^ 2 + y ^ 2 = 1, P is a moving point on the circle, the bisector of ∠ POA intersects Pa at Q, and the trajectory of Q point is obtained

Let Q (x, y) P on x2 + y2 = 1, P (COSA, Sina) angle bisector theorem AQ: QP = OA: OP = 3:1, x = (3 + 3cosa) / (1 + 3) y = (0 + 3sina) / (1 + 3) simplify to: (4y-3) square + (4Y) square = 9, that is: (Y-3 / 4) square + (y) square = 9 / 16, where x is not equal to 0

Given a (2,2), if P is a moving point on the circle x2 + y2 = 4, then the trajectory equation of the midpoint m of the line AP is______ ．

Let m (x, y) and P (m, n) be the midpoint of AP, then M2 + N2 = 4 & nbsp; ①. From the midpoint formula, we can get x = 2 + m2, y = 2 + N2, ｜ M = 2x-2, and N = 2y-2 ②. Substituting ② into ①, we can get 4 (x-1) 2 + 4 (Y-1) 2 = 4, that is, (x-1) 2 + (Y-1) 2 = 1, so the answer is: (x-1) 2 + (Y-1) 2 = 1