Given that the equation of a circle is X & sup2; + Y & sup2; - 4x + 4Y + 4 = 0, then the shortest distance from the point on the circle to the origin is and the longest distance is

Given that the equation of a circle is X & sup2; + Y & sup2; - 4x + 4Y + 4 = 0, then the shortest distance from the point on the circle to the origin is and the longest distance is


x²-4x+4+y²+4y+4=4
(x-2)²+(y+2)²=4
Represents a circle whose center (2, - 2) radius is 2
Combination of number and shape
The distance from the center of the circle to the origin = √ (2-0) &# 178; + (- 2-0) &# 178; = 2 √ 2
So the shortest distance = 2 √ 2-2, the longest distance = 2 + 2 √ 2



As shown in the figure, the intersection circle x2 + (Y-1) 2 = 1 of the moving line passing through the origin is at point Q. take point P on the line OQ so that the distance from P to the line y = 2 is equal to | PQ | and solve the trajectory equation of point P when the moving line rotates around the origin


P (x, y, P (x, y), O1: x2 + (Y-1) 2 = 1 and y = 2, circle O1: x2 + (Y-1) 2 = 1 and line y = 2, tangent to point a, connect AQ, easy to know |aq ? = |||||||||||aq | = |||\\\\124\124\\\| ar = \124\\124\\\\\\\\x2 + y 2 = 4 is the orbit equation