1. Draw a secant from a point outside the circle P (1,1) to the circle x2 + y2 = 1, intersect the circle at two points a and B, and find the trajectory equation of the midpoint of the chord ab

1. Draw a secant from a point outside the circle P (1,1) to the circle x2 + y2 = 1, intersect the circle at two points a and B, and find the trajectory equation of the midpoint of the chord ab

Let the slope be K
y-1=k(x-1)
y=kx+(1-k)
Substituting
(k²+1)x²+2k(1-k)x+(1-k)²-1=0
x1+x2=-2k(1-k)/(k²+1)
y=kx+(1-k)
So Y1 + y2 = K (x1 + x2) - 2 (1-k) = 2 (1-k) / (K & sup2; + 1)
Midpoint x = (x1 + x2) / 2, y = (Y1 + Y2) / 2
So x / y = - K
And Y-1 = K (x-1)
So k = (Y-1) / (x-1)
So x / y = - (Y-1) / (x-1)
-y²+y=x²-x
x²+y²-x-y=0
(x-1/2)²+(y-1/2)²=1/2
A straight line is a secant
The distance from the center of the circle (0,0) to the straight line is less than the radius
Straight line kx-y + (1-k) = 0
So | 1-k | / √ (K & sup2; + 1)