The trajectory of a point whose distance from the plane to the fixed point O is equal to 2cm is ()

A circle with a radius of 2cm centered at point o

If point P is outside plane a, then the locus of the point in plane a is equal to the distance from point P
What is the track.

The trajectory is a circle
Make a vertical line from this point to the plane. Take the perpendicular foot point as the center of the circle and the fixed length as the radius to draw a circle on the plane. Then the distance from all points on the circle to point P is equal

In plane rectangular coordinate system, the distance from point P to fixed point F (2,0) is equal to the distance to fixed line x + 2. If the locus of point P is curve C, the equation of curve C is obtained

Obviously, the curve is a parabola
At the same time, the distance from the origin to f (2,0) is equal to the distance to the fixed line x + 2, both are 2, so the vertex of the parabola is at the origin
So this parabola is a typical parabola with right opening, and its standard equation is y ^ 2 = 2px
Then let's see what P is here
According to the standard equation of parabola: in the parabola y ^ 2 = 2px, the focus is (P / 2,0), and the equation of Quasilinear is x = - P / 2
Compared with the known conditions, it is obvious that P / 2 = 2, then p = 4
So the equation of the curve is y ^ 2 = 4x

The equation for finding the locus of the moving point P whose distance from the fixed point F (1,0) is equal to the distance from the straight line x + 1 = 0

Let: P (x, y)
√[(x-1)²+y²]=x+1
Square: (x-1) & sup2; + Y & sup2; = (x + 1) & sup2;
x²-2x+1+y²=x²+2x+1
y²=4x

The trajectory of a point whose distance from the plane to the fixed point O is equal to 2cm is ()