Intersection beween circle and line

Question

So I have the circl $x^2+y^2=5$ and the line $y=2x+c$ and I want to find all the points were the two intersect. I know how to solve this for a specific value of $c$, but I do not know how to handle the problem for a general case. Any sugestions?

Answer 1

HINT

Substituting $y = 2x + c$ in $x^{2} + y^{2} = 5$ results $x^{2} + (2x + c)^{2} = 5.$ So, the equation is

$$5x^2+4cx+c^2-5=0.$$

The solution then is

$$x_{1,2}=\frac{-4c\pm\sqrt{16c^2-20(c^2-5)}}{10}.$$

That is, $c$ is simply a parameter of the solution.

If $$16c^2-20(c^2-5)<0$$ then there is no intersection.

Answer 2

$$ \begin{cases} x^2+y^2=5\\ y=2x+c \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x^2+(2x+c)^2=5\\ y=2x+c \end{cases}\Longleftrightarrow $$ $$ \begin{cases} 5x^2+4xc+c^2=5\\ y=2x+c \end{cases}\Longleftrightarrow $$ $$ \begin{cases} 5x^2+4xc+c^2-5=0\\ y=2x+c \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x=\frac{-4c\pm\sqrt{\left(4c\right)^2-4\cdot 5\cdot \left(c^2-5\right)}}{2\cdot 5}\\ y=2x+c \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x=\frac{1}{5}\left(\pm\sqrt{25-c^2}-2c\right)\\ y=2x+c \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x=\frac{1}{5}\left(\pm\sqrt{25-c^2}-2c\right)\\ y=2\left(\frac{1}{5}\left(\pm\sqrt{25-c^2}-2c\right)\right)+c \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x=\frac{1}{5}\left(\pm\sqrt{25-c^2}-2c\right)\\ y=\frac{1}{5}\left(c\pm 2\sqrt{25-c^2}\right) \end{cases} $$

So the solutions are:

$$x_1=\frac{1}{5}\left(-\sqrt{25-c^2}-2c\right)\space,\space y_1=\frac{1}{5}\left(c-2\sqrt{25-c^2}\right)$$ $$x_2=\frac{1}{5}\left(\sqrt{25-c^2}-2c\right)\space,\space y_2=\frac{1}{5}\left(c+2\sqrt{25-c^2}\right)$$