Interection of Lines - Create the missing equation

Question

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.

Answer 1

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.

Answer 2

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.