Given the point Q (2,0) and the circle O: X & # 178; + Y & # 178; = 1, the ratio of the tangent length from the moving point m to the circle to | MQ | is root 2, the trajectory equation of the moving point m is obtained For detailed explanation, please refer to the attached figure,

Given the point Q (2,0) and the circle O: X & # 178; + Y & # 178; = 1, the ratio of the tangent length from the moving point m to the circle to | MQ | is root 2, the trajectory equation of the moving point m is obtained For detailed explanation, please refer to the attached figure,


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, which is the root sign, the ratio of the tangent length to MQ is the root sign 2, and the root sign x ^ 2 + y ^ 1 / root sign (X-2) ^ 2 + y ^ 2 = root sign 2, which is x ^ 2 + y ^ 2



Given a point a (2,0) and a circle x ^ 2 + y ^ 2 = 1 on the rectangular coordinate plane, the ratio of the tangent length | MB | and | Ma | from the moving point m to o is the root sign 2, the trajectory% of the moving point m is obtained


Let m (x, y), | MB | ^ 2 = (x ^ 2 + y ^ 2) ^ 1 / 2-1, | Ma | = ((X-2) ^ 2 + y ^ 2) ^ 1 / 2,
|MB|/|MA|=2^1/2,
The trajectory of M is x ^ 2 + y ^ 2-8x + 9 = 0



The trajectory equation of M can be obtained from the point Q (2.0) and the circle x ^ 2 + y ^ 2 = 1 in the rectangular coordinate system, and the tangent length Mn from the moving point m to the circle is MQ
N is a point on a circle and M is the same as M


Let m (x, y), link on, OM, then in RT △ omn: according to Pythagorean theorem: | Mn | ^ 2 + | on | ^ 2 = | om | ^ 2, that is, | Mn | ^ 2 = (x ^ 2 + y ^ 2) - 1 | Mn | = √ (x ^ 2 + y ^ 2) - 1 ∵ Mn = MQ ∵ √ (x ^ 2 + y ^ 2) - 1 = √ (2-x) ^ 2 + y ^ 2 (x ^ 2 + y ^ 2) - 1 = (2-x) ^ 2 + y ^ 2 + y ^ 2-1 = 4-4x