The moving circle is inscribed with the fixed circle x square + y square - 4y-32 = 0, and passes through a fixed point a (0, - 2) in the circle to find the trajectory equation of the center of the moving circle Thank you,

The moving circle is inscribed with the fixed circle x square + y square - 4y-32 = 0, and passes through a fixed point a (0, - 2) in the circle to find the trajectory equation of the center of the moving circle Thank you,

If you draw a picture, you will find that the coordinates of the center of a circle are actually ellipses with focus on (0,2) and (0, - 2) and 2A = 6. What is required is 2B,
We can know that when the point y = 0, the distance from the point to the center of the great circle is 3, we can know that B ^ 2 = 3 ^ - 2 ^ = 5, so the trajectory equation is:
x^2/5+y^2/9=1.

A circle passes through two points a (4,2), B (- 1,3), and the sum of four intercepts on two coordinate axes is 2

Let the equation of circle be x2 + DX + Y2 + ey + F = 0, and substitute two points a (4, 2) and B (- 1, 3) into the equation to get: e = 5D + 10, f = - 14d-40, because the four intercepts are 2, so - D-E = 2, so the solution is: D = - 2, f = - 12, e = 0, so the equation of circle is x2-2x + y2-12 = 0, that is, (x-1) 2 + y2 = 13

A circle passes through two points a (4,2), B (- 1,3), and the sum of four intercepts on two coordinate axes is 2

Let the equation of circle be x2 + DX + Y2 + ey + F = 0, and substitute two points a (4, 2) and B (- 1, 3) into the equation to get: e = 5D + 10, f = - 14d-40, because the four intercepts are 2, so - D-E = 2, so the solution is: D = - 2, f = - 12, e = 0, so the equation of circle is x2-2x + y2-12 = 0, that is, (x-1) 2 + y2 = 13