Through a point m (a, b) outside the circle x ^ 2 + y ^ 2 = R ^ 2 to make the secant of the circle, the trajectory equation of the middle point of the chord is obtained

Through a point m (a, b) outside the circle x ^ 2 + y ^ 2 = R ^ 2 to make the secant of the circle, the trajectory equation of the middle point of the chord is obtained

AB midpoint P (x, y)
xA+xB=2x
yA+yB=2y
(xA)^2+(yA)^2=r^2.(1)
(xB)^2+(yB)=4y^2.(2)
(1)-(2):
(xA+xB)*(xA-xB)+(yA+yB)*(yA-yB)=0
(xA+xB) +( yA+yB)*(yA-yB)/(xA-xB)=0
k(AB)=(yA-yB)/(xA-xB)=(y-b)/(x-a)
2x+2y*(y-b)/(x-a)=0
The trajectory equation of midpoint AB is circle: x ^ 2 + y ^ 2-ax-by = 0

The trajectory equation of the midpoint of the chord of length 2 in the circle 〈 X-2 〉 ^ 2 + ^ 2 = 9
The trajectory equation of the midpoint of the chord of length 2 in the circle 〈 X-2 〉 ^ 2 + ^ 2 = 9

(x-2)^2+(y+1)^2=9
That is, the center coordinates o (2, - 1), radius r = 3
Let the coordinates of point m be: m (x, y)
|OM|^2=(X-2)^2+(y+1)^2
According to Pythagorean theorem, OM ^ 2 + (2 / 2) ^ 2 = R ^ 2
(x-2)^2+(y+1)^2+1=9
The midpoint equation is: (X-2) ^ 2 + (y + 1) ^ 2 = 8