It is known that the straight line X-Y + 3 = 0 and the circle x + y = 1. Judge their position relation and prove it

It is known that the straight line X-Y + 3 = 0 and the circle x + y = 1. Judge their position relation and prove it

The distance between the center of circle 0 and the straight line d = (0-0 + 3) / root sign (1 ^ 2 + 1 ^ 2) = 3 / root sign 2 > R = 1, so it is separated

(1) The equation of a circle is known as the square of X + the square of Y - 2x + 2Y = 0
(1) If the equation of a circle is known to be the square of X + the square of Y - 2x + 2Y = 0, the circumference of the circle can be obtained. (2) if the line L passes through the point (1,2) and is parallel to the line 3x + 4y-12 = 0, the equation of the line l can be obtained

(1)
x²+y²-2x+2y=0
(x-1)²+(y+1)²=2
So the radius is 2
(2) Let the linear equation be 3x + 4Y + C = 0 (/ / A1: A2 = B1: B2)
Passing point (1,2)
∴ 3+8+c=0
c=-11
The linear equation is 3x + 4y-11 = 0

It is known that the symmetric point of any point on C: x2 + Y2 + 2x + ay-3 = 0 (a is a real number) with respect to the line L: X-Y + 2 = 0 is on the circle C. then a=
Your answer was originally a = - 2, any point on the circle is symmetrical about the straight line, which means that the center of the circle on the straight line is known by x2 + Y2 + 2x + ay-3 = 0, and the center of the circle (- 1, y) is brought into X-Y + 2 = 0 to get y = 1, so the center of the circle (- 1, 1) is brought into the analytical formula of the circle, and a = - 2
But substituting (- 1,1), a = 3? Why

If any point of symmetry about the line L: X-Y + 2 = 0 on the circle C is on the circle C, then the line passes through the center of the circle, and a
It is known that the straight line X-Y + 2 = 0 passes through the center of the circle (1, - A / 2),
So &; 1 + A / 2 + 2 = 0,
So a = - 2
So a = - 2