\(c{:}\,(x + 6)^{2} + (y + 5)^{2} = 4 \text{.}\)

\(r = d(M , A) = \sqrt{(-3 - 1)^{2} + (-1 - -4)^{2}} = \sqrt{25} \text{.}\)

\(c{:}\,(x + 3)^{2} + (y + 1)^{2} = 25 \text{.}\)

Het middelpunt \(M\) is het midden van \(A\kern{-.8pt}B \text{,}\) dus
\(M ({1 \over 2} (2 + -4) , {1 \over 2} (1 + 0)) = M (-1 , \frac{1}{2}) \text{.}\)

\(r = d(M , A) = \sqrt{(-1 - 2)^{2} + (\frac{1}{2} - 1)^{2}} = \sqrt{9\frac{1}{4}} \text{.}\)

\(c{:}\,(x + 1)^{2} + (y - \frac{1}{2})^{2} = 9\frac{1}{4} \text{.}\)

Cirkel \(c\) raakt aan de \(x \text{-}\)as, dus \(r = d(M , x \text{-as}) = 1 \text{.}\)

\(c{:}\,(x - 2)^{2} + (y - 1)^{2} = 1 \text{.}\)

\(c{:}\,x^{2} + (y + 5)^{2} = 36 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 12 x - 2 y + 33 = 0\)
\((x + 6)^{2} - 36 + (y - 1)^{2} - 1 + 33 = 0\)
\((x + 6)^{2} + (y - 1)^{2} = 4 \text{.}\)

Dus \(M (-6 , 1)\) en \(r = \sqrt{4} = 2 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 3 x - 12 y + 21 = 0\)
\((x - 1\frac{1}{2})^{2} - 2\frac{1}{4} + (y - 6)^{2} - 36 + 21 = 0\)
\((x - 1\frac{1}{2})^{2} + (y - 6)^{2} = 17\frac{1}{4} \text{.}\)

Dus \(M (1\frac{1}{2} , 6)\) en \(r = \sqrt{17\frac{1}{4}} \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 12 y - 13 = 0\)
\(x^{2} + (y - 6)^{2} - 36 - 13 = 0\)
\(x^{2} + (y - 6)^{2} = 49 \text{.}\)

Dus \(M (0 , 6)\) en \(r = \sqrt{49} = 7 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,x - 3 y = c \\ A (-4 , -3)\end{rcases} c = 1 ⋅ -4 - 3 ⋅ -3 = 5\)
Dus \(n{:}\,x - 3 y = 5 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}3 x + y = -5 \\ x - 3 y = 5\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}3 x + y = -5 \\ 3 x - 9 y = 15\end{cases}\)
Aftrekken geeft \(10 y = -20\) dus \(y = -2 \text{.}\)

\(\begin{rcases}3 x + y = -5 \\ y = -2\end{rcases} \begin{matrix}3 x + 1 ⋅ -2 = -5 \\ x = -1\end{matrix}\)
Dus \(S (-1 , -2) \text{.}\)

\(d(A , l) = d(A , S) = \sqrt{(-4 - -1)^{2} + (-3 - -2)^{2}} = \sqrt{10} \text{.}\)

\(A (-4 , -3)\) en \(r = d(A , l) = \sqrt{10} \text{,}\) dus
\(c{:}\,(x + 4)^{2} + (y + 3)^{2} = 10 \text{.}\)

\(c{:}\,(x + 3)^{2} + (y + 6)^{2} = 16 \text{.}\)

Haakjes wegwerken geeft
\(x^{2} + 6 x + 9 + y^{2} + 12 y + 36 = 16\)
en dus
\(c{:}\,x^{2} + y^{2} + 6 x + 12 y + 29 = 0 \text{.}\)

De cirkels raken de \(x \text{-}\)as en hebben straal \(3 \text{,}\) dus \(y_{M} = 3\) of \(y_{M} = -3 \text{.}\)

\(\begin{rcases}y = x + 4 \\ y_{M} = 3\end{rcases} \text{ geeft } x + 4 = 3 \text{ dus } x_{M} = -1\)

Middelpunt \(M_{1} (-1 , 3)\) en straal \(r = 3 \text{,}\) dus
\(c_{1}{:}\,(x + 1)^{2} + (y - 3)^{2} = 9\)

Op dezelfde manier geldt dat
\(\begin{rcases}y = x + 4 \\ y_{M} = -3\end{rcases} \text{ geeft } x + 4 = -3 \text{ dus } x_{M} = -7\)
Middelpunt \(M_{2} (-7 , -3)\) en straal \(r = 3 \text{,}\) dus
\(c_{2}{:}\,(x + 7)^{2} + (y + 3)^{2} = 9\)

\(x = -3\) invullen in de vergelijking van \(c\) geeft
\((-3)^{2} + y^{2} + 8 ⋅ -3 + 6 y + 20 = 0 \text{.}\)

De vergelijking oplossen geeft
\(y^{2} + 6 y + 5 = 0\)
\((y + 5) (y + 1) = 0\)
\(y = -5 ∨ y = -1\)
\(y_{A} > y_{B} \text{,}\) dus \(A (-3 , -1)\) en \(B (-3 , -5) \text{.}\)