Given the circle m; 2x * x + 2Y * y-8x-8y-1 = 0 and the straight line L: x + Y-9 = 0, make a triangle ABC through a point a on the straight line L to make the angle BAC = 45 degrees, AB passes through the center m, and B, C are on the circle M 1: When the abscissa of a is 4, the equation of line AC is obtained 2: Find the value range of abscissa of point a

Given the circle m; 2x * x + 2Y * y-8x-8y-1 = 0 and the straight line L: x + Y-9 = 0, make a triangle ABC through a point a on the straight line L to make the angle BAC = 45 degrees, AB passes through the center m, and B, C are on the circle M 1: When the abscissa of a is 4, the equation of line AC is obtained 2: Find the value range of abscissa of point a

The equation of circle M can be rewritten as (X-2) ^ 2 + (Y-2) ^ 2 = (√ 34 / 2) ^ 2, which means that the center of circle m is O (2,2), and the radius r = √ 34 / 2. Point a is on the straight line L, so when abscissa x = 4, ordinate = 5, AB passes through the center O (2,2), so the slope of AB K1 = (5-2) / (4-2) = 3 / 2. (1) let the slope of AC K2, the angle between AB and AC be 45 ° = > t

Given that the circle M: 2x2 + 2y2-8x-8y-1 = 0, the straight line L: x + Y-9 = 0, make △ ABC through a point a on L, so that ∠ BAC = 45 °, the edge AB passes through the center m, and B, C are on the circle m, find the value range of the ordinate of point a

From 2x2 + 2y2-8x-8y-1 = 0, the standard equation of a circle is obtained as follows: (X-2) 2 + (Y-2) 2 = 172, the center of the circle m (2, 2), radius r = 342, ∵ straight line L: x + Y-9 = 0, ∵ let a (9-A, a), ∵ B, C be on the circle m, the intersection or tangency of the straight line AC and the circle m, the distance d from the center of the circle m to AC ≤ R, ? BAC = 45 degree, ? d = 22 | am | ≤ R, that is 22 · (7 − a) 2 + (a − 2) 2 ≤ 342 The solution is 3 ≤ a ≤ 6, so the range of the ordinate of point a is [3,6]

Given that the circle M: 2x2 + 2y2-8x-8y-1 = 0, the straight line L: x + Y-9 = 0, make △ ABC through a point a on L, so that ∠ BAC = 45 °, the edge AB passes through the center m, and B, C are on the circle m, find the value range of the ordinate of point a

From 2x2 + 2y2-8x-8y-1 = 0, the standard equation of a circle is obtained as follows: (X-2) 2 + (Y-2) 2 = 172, the center of the circle m (2, 2), radius r = 342, ∵ straight line L: x + Y-9 = 0, ∵ let a (9-A, a), ∵ B, C be on the circle m, the intersection or tangency of the straight line AC and the circle m, the distance d from the center of the circle m to AC ≤ R, ? BAC = 45 degree, ? d = 22 | am | ≤ R, that is 22 · (7 − a) 2 + (a − 2) 2 ≤ 342 The solution is 3 ≤ a ≤ 6, so the range of the ordinate of point a is [3,6]