To prove that y = 2x - 10 is tangent to the circle we find the center and radius of the circle and show that the distance of the center from the line is equal to the radius.
The circle: x^2 - 4x + y^2 + 2y = 0
=> x^2 - 4x + 4 + y^2 + 2y + 1 = -5
=> (x - 2)^2 + (y + 1)^2 = (sqrt 5)^2
The center of the circle is (2, -1) and the radius is sqrt 5
The distance of the point (2, -1) from y = 2x - 10 is
|2*2 + 1 - 10|/sqrt (4 + 1)
=> 5/sqrt 5
=> sqrt 5
We see that the point (2, -1) does lie at a distance equal to (sqrt 5) from the line 2x - y - 10 = 0
This proves that y = 2x - 10 is tangent to circle x^2 - 4x + y^2 + 2y = 0
To prove that the line is tangent to the given circle we'll have to solve the system formed by the equations of the line and circle and to demonstrate that it has only a single solution.
This solution of the system represents the point where the line is tangent to the circle.
y = 2x - 10 (1)
We'll substitute (1) 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
y = f(4) = -2
We notice that the system has a single solution, therefore the line is tangent to the circle in the point whose coordinates are (4;-2).