<s>[INST] How should I have derived the extra solution my book lists? I am working through some geometry exercises, one of which is to find the point(s) where the circle defined by $x^2 + y^2 = 16$ touches the line defined by $x - 2y = 4$.
I defined $x$ in terms of $y$ starting from the equation for the line: $x = 4 + 2y$.
Substituted in the equation for the circle this yields $(4 + 2y)^2 + y^2 = 16$. The abc-formula gives $y = \frac{-20 \pm 20}{2} \implies y = 0 \lor y = -20$. $-20$ can't be a solution since $x^2 + 400 = 16$ has no solutions, so one solution that satisfies both equations is $(4, 0)$.
Then I checked my answer to see that, although my answer was correct, my book listed another solution: $(-\frac{12}{5}, -\frac{16}{5})$. Substitution shows that this indeed satisfies both equations, yet I never got there. How should I have derived this second solution?
TL;DR
I only found one solution to the following system of equations - $(4,0)$ -, yet my book lists two:
$$
\left\{ 
\begin{array}{c}
x^2 + y^2 = 16 \\ 
x - 2y = 4
\end{array}
\right.
$$
How should I have derived the second solution - $(-\frac{12}{5}, -\frac{16}{5})$ -?
 [/INST] plugging $$x=4+2y$$ in $$x^2+y^2=16$$ we get $$(4+2y)^2+y^2=16$$ from here we get
$$16+4y^2+16y+y^2=16$$ or equivalent to $$y(5y+16)=0$$ can you proceed?
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