From fixed point a (3,0) to fixed circle x ^ 2 + y ^ 2 = 2, make a straight line intersecting circle at point P, and find the trajectory equation of the midpoint of line AP

Let this point be (x, y) because a (3,0), so p (2x-3,2y) because the point is the midpoint of AP
And because P satisfies the fixed circle x ^ 2 + y ^ 2 = 2, it brings in (2x-3) ^ 2 + (2Y) ^ 2 = 2
Just simplify it

The equation of a circle is x2 + (Y-1) 2 = 4, passing through point a (0,3) as the secant of the circle, intersecting the circle with point P, and finding the locus of the midpoint of the line AP

Let the midpoint of AP be q (x, y), P (x0, Y0). From the question meaning: q is the midpoint of AP, we can get: x = x0 / 2, y = (Y0 + 3) / 2. With a little deformation, we can get: x0 = 2x, Y0 = 2y-3

Given the fixed point (3,0), the point a moves on the circle x ^ + y ^ = 1, M is a point on the line AB, and the vector am = 1 / 3 vector MB, then the trajectory equation of the point is?
The answer is (x-3 / 4) ^ + y ^ = 9 / 16

To solve the problem of trajectory equation, the coordinates of points should be set, and the corresponding equations should be brought in according to the conditions of the problem
Let a (x1, Y1), m (X2, Y2), the equation of circle is: X1 ^ 2 + Y1 ^ 2 = 1,
Then the vector am = (x2-x1, y2-y1), MB = (3-x2, - Y2)
Because the vector am = 1 / 3 and the vector MB, there are: x2-x1 = 1 / 3 (3-x2) ①, y2-y1 = 1 / 3 (- Y2) ②
The results show that: X1 = 4 / 3x2-1, Y1 = 4 / 3y2
Substituting X1 ^ 2 + Y1 ^ 2 = 1, we can get: (4 / 3x2-1) ^ 2 + (4 / 3y2) ^ 2 = 1
It is concluded that: (x2-3 / 4) ^ 2 + Y2 ^ 2 = 9 / 16
That is, the trajectory equation of point m is: (x-3 / 4) ^ 2 + y ^ 2 = 9 / 16
I hope you can give me some points,

From fixed point a (3,0) to fixed circle x ^ 2 + y ^ 2 = 2, make a straight line intersecting circle at point P, and find the trajectory equation of the midpoint of line AP