Circle x ^ + y ^ = 1 and circle x ^ + y ^ - 2x-2y = 0, the position relation is? (^ is 2, that is, x square or Y Square) So R + r = 1 + radical 2, but I calculate that the distance between the two center points is = radical 2. Why is it tangent??? The distance between the center of the circle is less than the sum of the radii, so it intersects!!!!!

Circle x ^ + y ^ = 1 and circle x ^ + y ^ - 2x-2y = 0, the position relation is? (^ is 2, that is, x square or Y Square) So R + r = 1 + radical 2, but I calculate that the distance between the two center points is = radical 2. Why is it tangent??? The distance between the center of the circle is less than the sum of the radii, so it intersects!!!!!

x^+y^-2x-2y=0,
The results show that (x-1) ^ + (Y-1) ^ = 2,
The square of the center of the circle in radius (1,1) is 2
It is tangent to the circle whose center is at the origin and radius is 1
The center of a circle is the origin, the center of a circle is (1,1), and the distance between the centers is the root sign 2
The distance between the center of the circle is less than the sum of the radii, so it is tangent

It is known that circle C: x square + y square - 4 = 0. Line L: mx-y + 1-m = 0 (1) judge the position relationship between line L and circle C (2) if line L and circle C intersect at two different points

1. The straight line passes through the fixed point (1,1), which is in the circle, so the straight line intersects the circle
2. The vertical diameter theorem can be considered as long as the distance from the center of the circle to the straight line is less than the radius