Step-by-step explanation:

The equation of the given circle, x² + y² = 5, is in standard form and tells us that the center of the circle is at the origin, i.e., at the point (0, 0), and has a radius of r = √5, and the equation of the given line is y = 2x ‒ k is the slope-intercept form for the equation of a straight line, i.e., y = mx + b, where slope m = 2 and y-intercept b = ‒k.

As the first step in finding the desired, exact values of “k” for which the given line y = 2x ‒ k is tangent to the given circle, we’ll consider the equation of the given circle and solve for y in terms of x as follows:

x² + y² = 5

y² + x² = 5

y² + x² ‒ x² = 5 ‒ x²

y² = 5 ‒ x²

Now, taking the square root of both sides, we have:

y = ±√(5 ‒ x²)

Consequently, we have:

y = √(5 ‒ x²) , a semicircle, is the HALF of the given circle above the x-axis, and …

y = ‒√(5 ‒ x²), a semicircle, is the other half of the given circle below the x-axis.

At the point of tangency or intersection, (x, y), between the given circle and the given line, the circle and the line have a point in common; therefore, at that point, their x-values or coordinates and their y-values or coordinates will be the same, i.e., equal; consequently, we can set the two y-values equal to each other and solve for “k” as follows:

(1.) For the case of semicircle y = √(5 ‒ x²):

y = y (Equality is reflexive, i.e., for any real number a, a = a)

√(5‒ x²) = 2x ‒ k

√(5‒ x²) + k = 2x ‒ k + k

√(5‒ x²) + k = 2x + 0

√(5‒ x²) + [‒√(5 ‒ x²)] + k = 2x + [‒√(5 ‒ x²)]

0 + k = 2x + [‒√(5 ‒ x²)]

k = 2x + [‒√(5 ‒ x²)]

k = 2x ‒√(5 ‒ x²) is the generic VALUE of k for which the given line y = 2x ‒ k is tangent to the given circle x² + y² = 5 above the x-axis.

(2.) For the case of the other semicircle: y = ‒√(5 ‒ x²):

y = y (Equality is reflexive, i.e., for any real number a, a = a)

‒√(5 ‒ x²) = 2x ‒ k

‒√(5 ‒ x²) + k = 2x ‒ k + k

‒√(5 ‒ x²) + k = 2x + 0

‒√(5 ‒ x²) + √(5 ‒ x²) + k = 2x + √(5 ‒ x²)

0 + k = 2x + √(5 ‒ x²)

k = 2x + √(5 ‒ x²) is the generic value of k for which the given line y = 2x ‒ k is tangent to the given circle x² + y² = 5 below the x-axis.

We now have the two generic values of k: k = 2x ‒√(5 ‒ x²) and k = 2x + √(5 ‒ x²). These two generic values of k correctly imply two tangent lines and, consequently, two points of tangency with the given circle x² + y² = 5, i.e., we actually have two tangent lines, not just one; however, in order to determine the exact values of “k”, we have to know the x-coordinates of the two points of tangency since the two generic values of “k” are in terms of the x-coordinates of the points of tangency.

We know that the slope of each of the two tangent lines is m = 2 since both are represented by the given equation: y = 2x ‒ k. We also know that the slope of a tangent line to a curve is equal to the slope of the curve at the point of tangency (x, y), and the slope of the curve at the point of tangency is equal to the value of its function’s first derivative at that point, i.e., the slope m of the curve at the point of tangency (x, y) = dy/dx. (NOTE: Even though the given circle x² + y² = 5 does not represent a function, implicitly it represents two functions, that is, each of its two semicircles: y = √(5 ‒ x²) and y = ‒√(5 ‒ x²) explicitly do represent functions!)

So, to find the x-coordinates of the two points of tangency, (x, y), we need to first USE IMPLICIT differentiation to differentiate the implicit equation of the given circle, x² + y² = 5, with respect to x in order to give us a formula for dy/dx which is equal to the slope of the circle at any point on the circle (except on the x-axis), in particular, at the two points of tangency: