Tangents.Find if the line 2x-y-10=0 is tangent to the circle x^2+y^2-4x+2y=0.
If the line (2x - y - 10 = 0) is tangent to x^2+y^2-4x+2y=0, the two touch each other only at one point.
2x - y - 10 = 0
=> y = 2x - 10
Substituting in x^2+y^2-4x+2y=0
=> 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) = 0
=> x = 4
The point of contact is only (4, -2), therefore the line 2x-y-10=0 is a tangent to the circle x^2+y^2-4x+2y=0
The constraint that a line is tangent to a circle is given by the system formed by the equations of the line and circle that has to have one solution.
We'll change the equation of the line:
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.