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Find the values of a for which the graphs $$ y=x+a \quad \text{and}\quad x^2 + y^2 = 9$$ intersect at $1$, $2$ and $0$ points

I had gotten up to the point where for the first one, where there is just one intersection, that I could use perfect squares, however that lead to an unsolvable equation (at least I thought it was) after i substituted in the other equation. I am seriously stuck with this hard intersection problem and i would greatly appreciate help.

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Hint

Find $a$ s.t. $$x^2+(x+a)^2=9\iff 2x^2+2ax+a^2-9=0$$ has $0$, $1$ or $2$ solution... it's a very elementary quadric equation.

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    I am very sorry, but the issue is that I do not follow after this step. How does one continue? I realise that this could very well be a simple quadratic, but are they simplified further to find a?? Thanks – 2017-02-13
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    Let denote $\Delta =-4a^2+72$ (the discriminant). For which $a$ do we have $\Delta >0$, $\Delta =0$ or $\Delta <0$ ? Then conclude. – 2017-02-13
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    Yes, thank you for that, however, how does one get the discriminate (b^2-4ac) from this equation : 2x^2 + 2ax + a^2 - 9 = 0 ? The fact that it is a 4 - term quadratic is confusing me :( – 2017-02-13
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    For $\alpha x^2+\beta x+\gamma $, the discriminate is given by $\beta ^2-4\alpha \gamma $. Here $\alpha =2$, $\beta =2a$ and $\gamma =a^2-9$ – 2017-02-13
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    ah woops, im so stupid THANK YOU – 2017-02-13
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The first equation is a line, the second a circle around the origin with radius $r=3$.

So you have the cases:

  • no solution: line more than $3$ units away from the origin
  • one solution: line $3$ units away from the origin
  • two solution: line less than $3$ units away from the origin

You can rewrite the line equation as: $$ y = x + a \iff \\ (-1, 1) (x,y)^T = a \iff \\ \underbrace{\frac{1}{\sqrt 2} (-1, 1)}_n (x,y)^Z = \frac{a}{\sqrt 2} =: d $$ where $n$ is a unit normal vector of the line (there is $-n$ too) and $d$ is the (signed) distance of the line to the origin.

This should allow you to pick an $a$ value for each case.