∫ The straight line L passes through the point (-2,0), so the analytical formula of the straight line L can be written as:

K(x+2)-y=0 is the general formula: kx-y+2k=0.

∵ Circle X again? +y？ = 1 tangent, so the distance from the center of the circle to the straight line is equal to the radius.

The coordinate of the center of the circle is (0,0), so the distance from the center of the circle to the straight line is |2k|/ root sign (1+k 2).

The radius is 1. Combining the above formulas, we can get: k= root sign 3/3 or-root sign 3/3.

2. This problem can be solved by parameter method.

Let the coordinates of two points on the circle be (root number 3sina, root number 3cosa).

Then the distance from this point to the straight line x-y=3 is |3+ root 3cosa- root 3sina|/ root 2.

Using the auxiliary angle formula, the maximum distance can be obtained as: (the square root of 3+6)/root number 2.

3. The key is to determine the center and radius of the circle.

The radius is easy to find, r = (root number 5 of 3)/2.

As for the center coordinate, it can be solved by tangent length theorem.

Solution, the center coordinates are: (3- radical number 5)/2, (3- radical number 5)/2.

So the inscribed circle equation can be solved.

4. Because the straight line ax-by+c=0 and the circle X? +y？ = 1 tangency

So the distance from the center of the circle (0,0) to the straight line ax-by+c=0 is equal to the radius 1.

Using the distance formula from point to straight line, we can get |c|/ root sign (A 2+B 2) = 1.

So the solution is: A 2+B 2 = C 2.

So a triangle with three sides is a right triangle.

Choose B.

If you are satisfied, adopt me.