It is known that the radius of the center (0,2) of the circle m is 1, q is the moving point on the x-axis, QA and QB are tangent to the circle m, a and B respectively Finding the trajectory equation of the midpoint P of the moving string ab It's on the Internet, but I can't understand it, Well, 50 more points!

It is known that the radius of the center (0,2) of the circle m is 1, q is the moving point on the x-axis, QA and QB are tangent to the circle m, a and B respectively Finding the trajectory equation of the midpoint P of the moving string ab It's on the Internet, but I can't understand it, Well, 50 more points!

The simplest way to solve this problem is to use geometry
From the vertical diameter theorem, we know that MPQ is collinear and MQ ⊥ ab
It is shown that ∠ MAQ = 90 ° = ∠ MPa, so | MP | * | MQ | = | Ma | &# 178; = 1
Take the fixed point C (0,3 / 2), then C is on the line segment OM, and | MC | * | Mo | = 1 = | MP | * | MQ |
So △ MPC ∽ MOQ, thus ∠ MPC = ∠ MOQ = 90 °
Therefore, point P is located on a circle with diameter MC
Note that P cannot reach point a (otherwise the tangent lines PA and Pb are parallel)
So the equation is: X & # 178; + (Y-7 / 4) &# 178; = 1 / 16, y ≠ 2