# Tangents.Find if the line 2x-y-10=0 is tangent to the circle x^2+y^2-4x+2y=0.

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### 2 Answers

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:

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.**