CE amaths problem

Let P = (s, t)
Since P lies on the circle
So
s^2 + t^2 + 2s - 2t - 7 = 0 ..............(1)

Let M = (x, y)
Since Mis the mid-point of AP
So

x = (2 + s) /2
2x = 2 + s
s = 2x - 2 ..................(2)

Also
y = (3 + t) /2
2y = 3 + t
t = 2y - 3 ...................(3)

Put (2) and (3) into (1), we get
(2x - 2)^2 + (2y - 3)^2 + 2 (2x - 2) - 2 (2y - 3) - 7 = 0
4x^2 - 8x + 4 + 4y^2 - 12y + 9 + 4x - 4 - 4y + 6 - 7 = 0
4x^2 + 4y^2 - 4x - 16y + 8 = 0
x^2 + y^2 - x - 4y + 2 = 0

Equation of locus of point M is

x^2 + y^2 - x - 4y + 2 = 0

Let P = (s, t)
Since P lies on the circle
So
s^2 + t^2 + 2s - 2t - 7 = 0 ..............(1)

Let M = (x, y)
Since Mis the mid-point of AP
So

x = (2 + s) /2
2x = 2 + s
s = 2x - 2 ..................(2)

Also
y = (3 + t) /2
2y = 3 + t
t = 2y - 3 ...................(3)

Put (2) and (3) into (1), we get
(2x - 2)^2 + (2y - 3)^2 + 2 (2x - 2) - 2 (2y - 3) - 7 = 0
4x^2 - 8x + 4 + 4y^2 - 12y + 9 + 4x - 4 - 4y + 6 - 7 = 0
4x^2 + 4y^2 - 4x - 16y + 12 = 0
x^2 + y^2 - x - 4y + 3 = 0

Equation of locus of point M is

x^2 + y^2 - x - 4y + 3 = 0

Let M(x, y).
Let P(a, b).

x = (a + 2)/2
2x = a + 2
a = 2x - 2 --- (1)

y = (b + 3)/2
b = 2y - 3 --- (2)

Since P is on the circle,
a^2+b^2+2a-2b-7=0
(a + 1)^2 + (b - 1)^2 = 9 --- (3)
Put (1), (2) into (3),
(2x - 2 + 1)^2 + (2y - 3 - 1)^2 = 9
4(x - 1/2)^2 + 4(y - 2)^2 = 9.
(x - 1/2)^2 + (y - 2)^2 = 9/4