Given the point Q (2,0) and the circle C: x2 + y2 = 1 on the rectangular coordinate plane, when the ratio of the tangent length from the moving point m to the circle and MQ is 1 or 2 respectively, the trajectory equation of the point m is obtained

Given the point Q (2,0) and the circle C: x2 + y2 = 1 on the rectangular coordinate plane, when the ratio of the tangent length from the moving point m to the circle and MQ is 1 or 2 respectively, the trajectory equation of the point m is obtained

Let the coordinates of M be (x, y),
|MC|^2=x^2+y^2
r^2=1
Let the tangent length from the moving point m to the circle be d
d^2=|MC|^2-r^2=x^2+y^2-1
|MQ|^2=(x-2)^2+y^2
When D / MQ = 1, d = MQ, that is, d ^ 2 = | MQ | ^ 2
Then: x ^ 2 + y ^ 2-1 = (X-2) ^ 2 + y ^ 2
It is reduced to 4x = 5
x=5/4
When D / MQ = 2, d = 2mq, that is, d ^ 2 = 4 | MQ | ^ 2
Then: x ^ 2 + y ^ 2-1 = 4 [(X-2) ^ 2 + y ^ 2]
It is reduced to: 3Y ^ 2 + 3x ^ 2-16x + 17 = 0
(x-8/3)^2+y^2=13/9
As mentioned above:
When D / MQ = 1, the trajectory equation of point m is: x = 5 / 4
When D / MQ = 2, the trajectory equation of point m is: (X-8 / 3) ^ 2 + y ^ 2 = 13 / 9

Given the point Q (2,0) and the circle O: x ^ 2 + y ^ 2 = 1, the ratio of the tangent length from the moving point m to the circle O to | MQ | is root 2, the trajectory equation of the moving point m is obtained

Let m (x, y) be the tangent length from m to circle O. according to the Pythagorean theorem, subtract the radius of the circle from the square of the distance from the origin to M. then under the open root sign is the root sign, the ratio of the tangent length to MQ is the root sign 2, and the root sign x ^ 2 + y ^ 2-1 / root sign (X-2) ^ 2 + y ^ 2 = root sign 2, that is, x ^ 2 + y ^ 2-8x + 9 = 0. That is the trajectory equation of the moving point M!

The distance between the point m and the point F (4, O) is 1 less than the distance from the point m to the line L: x + 5 = o

Let m (x, y)
The distance between point m and point F (4,0)
=√[(x-4)^2+y^2]
The distance from it to the line L: x + 5 = O
=|x+5|
Determinant
√[(x-4)^2+y^2]+1=|x+5|
Simplification
y^2=16x
It's a parabola
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The distance between the point m and the point F (4,0) is 1 less than the distance from the point m to the line L: x + 5 = 0
Return information

The coordinates of solution point m are (x, y)
√[(x-4)^2+y^2]=(x+5)-1
(x-4)^2+y^2=(x+4)^2
y^2=16x
The trajectory equation of m point y ^ 2 = 16x