Find the value of k for which the line y = -x + k is a tangent to the circle (x+1)2 + (y−1)2 = 1.

The equation of the circle centred at M is $(x+1)^2 + (y-1)^2 = 1$.
The equation of the straight line is $y = -x + k$.
Substitute $y = -x + k$ into the circle equation:

$$(x+1)^2 + ((-x+k)-1)^2 = 1$$ $$(x+1)^2 + (-x + (k-1))^2 = 1$$

Expand the squared terms:

$$(x^2 + 2x + 1) + ((-x)^2 + 2(-x)(k-1) + (k-1)^2) = 1$$ $$x^2 + 2x + 1 + x^2 - 2x(k-1) + (k^2 - 2k + 1) = 1$$ $$x^2 + 2x + 1 - 2kx + 2x + k^2 - 2k + 1 = 1$$

Combine like terms and move the constant to the left side to set the equation to 0:

$$2x^2 + (2 - 2k + 2)x + (1 + k^2 - 2k + 1 - 1) = 0$$ $$2x^2 + (4 - 2k)x + (k^2 - 2k + 1) = 0$$

This shows that the equation can be written as $2x^2 + (4 - 2k)x + (k^2 - 2k + 1) = 0$.

For the line to be a tangent to the circle, the quadratic equation $2x^2 + (4 - 2k)x + (k^2 - 2k + 1) = 0$ must have exactly one real solution. This means its discriminant ($\Delta$) must be equal to zero.
The discriminant for a quadratic equation $ax^2 + bx + c = 0$ is $\Delta = b^2 - 4ac$.
From our equation, we have:
$a = 2$
$b = (4 - 2k)$
$c = (k^2 - 2k + 1)$

Set the discriminant to zero:

$$\Delta = (4 - 2k)^2 - 4(2)(k^2 - 2k + 1) = 0$$ $$(16 - 16k + 4k^2) - 8(k^2 - 2k + 1) = 0$$ $$16 - 16k + 4k^2 - 8k^2 + 16k - 8 = 0$$

Combine like terms:

$$(4k^2 - 8k^2) + (-16k + 16k) + (16 - 8) = 0$$ $$-4k^2 + 8 = 0$$ $$4k^2 = 8$$ $$k^2 = \frac{8}{4}$$ $$k^2 = 2$$ $$k = \pm\sqrt{2}$$

The numerical values of k are $\pm\sqrt{2}$.