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Find all values $a$ for which the system of equations it has exactly two solutions $$\begin{cases} x^2-8x+y^2-8(|x|+|y|-2)=4; \\ y=ax-4a-6 \end{cases}$$

My attempt:

1) $$x^2-8x+y^2-8(|x|+|y|-2)=4$$

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2) $\forall a \in \mathbb R$ $$(4;-6) \in y=ax-4a-6$$

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    I think that the solutions are: $a<-2$ and $a>{-10+\sqrt{28}\over3}$ – 2017-01-07

1 Answers 1

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As you wrote, $L:y=ax-4a-6$ is a line passing through $P(4,-6)$.

For $x\lt 0$, we have $$x^2-8x+y^2-8(-x+|y|-2)=4\iff x^2+(|y|-4)^2=4$$ For $x\ge 0$, we have $$x^2-8x+y^2-8(x+|y|-2)=4\iff (x-8)^2+(|y|-4)^2=68$$ These are the parts of four circles $$C_1 : x^2+(y-4)^2=4\ \ (x\lt 0,y\ge 0),\quad C_2 : x^2+(y+4)^2=4\ \ (x\lt 0,y\lt 0)$$ $$C_3 : (x-8)^2+(y-4)^2=68\ \ (x\ge 0,y\ge 0),\quad C_4 : (x-8)^2+(y+4)^2=68\ \ (x\ge 0,y\lt 0)$$

$\qquad\qquad$enter image description here

  • $C_1$ and $C_3$ intersect at $A(0,2)$ and $B(0,6)$.

  • $C_3$ and $C_4$ intersect at $C(8-2\sqrt{13},0)$ and $D(8+2\sqrt{13},0)$.

  • $C_2$ and $C_4$ intersect at $E(0,-2)$ and $F(0,-6)$.

  • The line $L:y=ax-4a-6$ is tangent to $C_1$ at $G$ when $a=\frac{-10+2\sqrt 7}{3}$.

  • The slope of the line $l$ passing through $P,F$ is $0$.

  • The slope of the line $g$ passing through $P,D$ is $\frac{\sqrt{13}-2}{3}$.

  • The slope of the line $h$ passing through $P,B$ is $-3$.

  • The slope of the line $i$ passing through $P,A$ is $-2$.

  • The slope of the line $j$ passing through $P,C$ is $\frac{-2-\sqrt{13}}{3}$.

  • The slope of the line $n$ passing $P,G$ is $\frac{-10+2\sqrt 7}{3}$.

  • The slope of the line $k$ passing through $P,E$ is $-1$.

We see that for $-2\le a\lt\frac{-10+2\sqrt 7}{3}$, the line $L$ and $C_1$ has two intersection points, so these $a$ have to be eliminated.

Also, for $a=\frac{-10+2\sqrt 7}{3}$, there are three intersection points, so this $a$ has to be eliminated.

Therefore, the answer is $$\color{red}{a\lt -2\quad\text{or}\quad a\gt \frac{-10+2\sqrt 7}{3}}$$