Functions and Equations - Values of k for which $y=2x-k$ is tangent to $x^{2}+y^{2}=5$

Question

The question asks: Find the values of k for which the line

$y=2x-k$ is tangent to the circle with equation $x^2+y^2=5$

So I started by substituting,

$x^2+(2x-k)^2=5$

$x^2+4x^2-4xk+k^2=5$

$5x^2+k^2-4xk-5=0$

But after this I couldn't see a way to factor it that would make able to find the discriminant and set that equal to $0$.

Help would be appreciated.

Answer 1

The distance from the point $(0,0)$ to the line $2x-y-k=0$ should be equal to the radius of the circle.

Thus, $$\frac{|2\cdot0-0-k|}{\sqrt{2^2+1^2}}=\sqrt5$$ or $$|k|=5.$$

Answer 2

Rewrite it as $$5x^2-(4k)x+(k^2-5)=0$$

Thus, $$\Delta=16k^2-20(k^2-5)=100-4k^2$$

Setting $\Delta=0$, we obtain $k=\pm 5$.

Visualization by Desmos:

Answer 3

Here is how I did it:

Differentiating $x^2+y^2=5$ gives $2x+2y\;dy/dx=0$, or $dy/dx=-x/y$.

The slope of the line $y=2x-k$ is $2$.

So at the tangent points, $dy/dx=-x/y=2$, i.e. $x=-2y$.

Plugging this back into the equation for the circle, we get $$(-2y)^2+y^2=5$$ $$y=\pm1$$ So the tangent points are $(2,-1)$ and $(-2,1)$. Substituting these into the equation for the line gives $k=\pm5$.