If the moving circle passes through the point (1,0) and is tangent to the straight line x = - 1, then the trajectory equation of the center of the moving circle? Ask for detailed explanation

If the moving circle passes through the point (1,0) and is tangent to the straight line x = - 1, then the trajectory equation of the center of the moving circle? Ask for detailed explanation

The distance from the circle to the point (1,0) is equal to the distance to the straight line x = - 1!
Let the center coordinates of the moving circle be (x, y), then
(x-1)^2+(y-0)^2=[x-(-1)]^2
That is, (x-1) ^ 2 + y ^ 2 = (x + 1) ^ 2 is reduced to y ^ 2 = 4x
It is a typical parabola

If the moving circle m passes through the point F (0,1) and is tangent to the straight line y = - 1, then the trajectory equation of the center of the moving circle is______ ．

Let the coordinates of the center of the moving circle be (x, y) ∵ the moving circle passes through the fixed point P (0, 1), and is tangent to the fixed line L: y = - 1, ∵ the distance from the center of the circle to the fixed point P and to the straight line L is equal to the radius, ∵ according to the definition of parabola, the trajectory equation of the center of the moving circle is x2 = 4Y, so the answer is x2 = 4Y

If the moving circle P is tangent to the fixed circle C: (x + 3) ^ 2 + y ^ 2 = 1 and tangent to the straight line L: x = 2, the trajectory equation of the moving circle center P is obtained
thank you

Let P (x, y)
Fixed circle C: (x + 3) ^ 2 + y ^ 2 = 1
Center C (- 3,0), radius 1
Let R be the radius of the moving circle, and draw a graph to show that the center of the circle is on the left side of x = 2
Then | PC | = R + 1, and | R | = 2-x
∴ |PC|=2-x+1=3-x
∴ |PC|²=(3-x)²
∴ (x+3)²+y²=(3-x)²
∴ x²+6x+9+y²=9-6x+x²
∴ y²=-12x
The trajectory equation of the moving circle center p y & # 178; = - 12x