Given that circle B: (x + 1) ^ 2 + y ^ 2 = 16 and point a (1,0), C is any point on the circle, the trajectory equation of the intersection P of AC vertical bisector L and segment CB is obtained It is better to set P (x, y) as any point on the trajectory

∵ P is on the vertical bisector of AC, so PA = PC, so Pb + PA = Pb + PC = radius of circle = 4, so the trajectory of P is an ellipse with a and B as the focus, 2A = 4, so the trajectory equation is not difficult

Given that circle B: (x + 1) 2 + y2 = 16 and point a (1,0), C is any point on the circle, the trajectory equation of the intersection point P of AC vertical bisector L and segment CB is obtained

&Because P is on the vertical bisector of AC, BP + PA = BP + PC = bcbc is the radius of ⊙ B, BC = 4, so the sum of distances from PA + Pb = 4P to a and B is a fixed value, and the trajectory is ellipse. The focus of ellipse is a and B, and the center is the midpoint of ab. because B (- 1,0), a (1,0), the middle point of AB is O (0,0), that is, the middle point of ellipse

Given the circle (x + 1) ^ 2 + y ^ 2 = 16, the center of the circle is B, the point a (1,0), and C is any point on the circle B, find the trajectory of the intersection P of AC vertical bisector and line segment CB
Detailed explanation

The distance from the focus to a is the distance to point C. Therefore, the sum of the distances from the focus to a and B is the radius 4 of the circle, which is an ellipse C = 12 A = 4 a = 2 A ^ 2 = 4 C ^ 2 = 1 B ^ 2 = 4-1 = 3
So the elliptic equation is x ^ 2 / 4 + y ^ 2 / 3 = 1

Given that circle B: (x + 1) 2 + y2 = 16 and point a (1,0), C is any point on circle B, the trajectory equation of the intersection P of AC vertical bisector L and segment CB is obtained

Let the coordinates of point C be (x, y). Of course, it satisfies the equation of circle
Let the midpoint of AC be D, and the coordinates can be known as ((x-1) / 2, Y / 2),
So the slope of the vertical line of AC is k = - (x-1) / y,
If you have a slope, if you have a point, you can write the equation of the vertical in AC
Then combine this equation with BC's equation (also expressed in X and y) to get the equation you want

Given that circle B: (x + 1) ^ 2 + y ^ 2 = 16 and point a (1,0), C is any point on the circle, the trajectory equation of the intersection P of AC vertical bisector L and segment CB is obtained It is better to set P (x, y) as any point on the trajectory