We know that 2x - y - 10 = 0 is tangential to x^2+y^2-4x+2y=0 if there is only one point of contact.

Now 2x - y - 10 = 0

=> y = 2x - 10

Substituting this in x^2+y^2-4x+2y=0, we get

x^2+ ( 2x - 10)^2 - 4x + 2( 2x - 10)=0

=> x^2 + 4x^2 + 100 - 40x - 4x + 4x - 20 =0

=> 5x^2 - 40x + 80 =0

=> x^2 - 8x + 16 =0

=> (x - 4)^2 = 0

=> x = 4

The point of contact is (4 , -2)

**Therefore 2x - y - 10 = 0 is tangential to x^2+y^2-4x+2y=0.**

The line is tangent to the circle if the system formed by the equatins of the line and circle has a solution.

We'll change the equation of the line:

2x-y-10=0

We'll isolate y to the left side:

-y = -2x + 10

y = 2x - 10

We'll substitute y in the equation of the circle:

x^2 + y^2 - 4x + 2y = 0

x^2 + (2x - 10)^2 - 4x + 2(2x - 10) = 0

We'll expand the square and remove the brackets:

x^2 + 4x^2 - 40x + 100 - 4x + 4x - 20 = 0

5x^2 - 40x + 80 = 0

We'll divide by 5:

x^2 - 8x + 16 = 0

We'll write the quadratic above as a perfect square:

(x - 4)^2 = 0

x1 = x2 = 4

**The line is tangent to the circle in the point whose coordinates are x = 4 and y = f(4) = -2.**

To verify if 2x-y-10 is a tangent to x^2+y^2-4x+2y=0.

We know that if 2x-y-10 is a tangent to the circle x^2+y^2-4x+2y=0, then the radius of the circle is equal to the distance of the centre C(h,k).

The given circle is written as: (x-2)^2-2^2+(y+1)^2-1^2 = 0

(x-2)^2+(y+1)^2 = 5.

So the centre is C(2, -1) and radius r= sqrt 5, or r^2 = 5.

Therefore the distance d of C(2, -1) from line 2x-y-10 = 0 is given by:

d= |2*xC- yC -10|/sqrt{2^2+(-1)^2} = |2*2-1*-1-10|/sqrt(5)

d = |5-10|/sqrt5 = sqrt5.

Therefore d = r = sqrt5 .

Therefore the given line is the tangent to the circle.