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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 - $(-\frac{12}{5}, -\frac{16}{5})$ -?

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    $\begingroup$ I'm voting to close this question as off-topic because since I made a very simple mistake (forgot to square one of the terms halfway solving an equation) which I missed again when I reviewed my work before posting. Some other SE sites have "simple typographical error" as close reason. $\endgroup$
    – 11684
    Commented Sep 1, 2016 at 14:43
  • $\begingroup$ The answerer who pointed out I must have made a mistake applying the quadratic formula unfortunately deleted their answer. $\endgroup$
    – 11684
    Commented Sep 1, 2016 at 14:45

1 Answer 1

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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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    $\begingroup$ I don't know why somebody downvoted this. I think it's a great answer since $y(5y + 16)$ makes sense of this seemingly random fraction $-\frac{16}{5}$. $\endgroup$
    – 11684
    Commented Sep 1, 2016 at 14:35

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