\(c{:}\,(x + 2)^{2} + (y + 6)^{2} = 49 \text{.}\)

\(r = d(M , A) = \sqrt{(-2 - 2)^{2} + (-1 - 1)^{2}} = \sqrt{20} \text{.}\)

\(c{:}\,(x + 2)^{2} + (y + 1)^{2} = 20 \text{.}\)

Het middelpunt \(M\) is het midden van \(A\kern{-.8pt}B \text{,}\) dus
\(M ({1 \over 2} (5 + 6) , {1 \over 2} (1 + 0)) = M (5\frac{1}{2} , \frac{1}{2}) \text{.}\)

\(r = d(M , A) = \sqrt{(5\frac{1}{2} - 5)^{2} + (\frac{1}{2} - 1)^{2}} = \sqrt{\frac{1}{2}} \text{.}\)

\(c{:}\,(x - 5\frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{1}{2} \text{.}\)

Cirkel \(c\) raakt aan de \(y \text{-}\)as, dus \(r = d(M , y \text{-as}) = 3 \text{.}\)

\(c{:}\,(x - 3)^{2} + (y - 1)^{2} = 9 \text{.}\)

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

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 4 x - 14 y + 17 = 0\)
\((x + 2)^{2} - 4 + (y - 7)^{2} - 49 + 17 = 0\)
\((x + 2)^{2} + (y - 7)^{2} = 36 \text{.}\)

Dus \(M (-2 , 7)\) en \(r = \sqrt{36} = 6 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 2 x + 11 y + 17 = 0\)
\((x - 1)^{2} - 1 + (y + 5\frac{1}{2})^{2} - 30\frac{1}{4} + 17 = 0\)
\((x - 1)^{2} + (y + 5\frac{1}{2})^{2} = 14\frac{1}{4} \text{.}\)

Dus \(M (1 , -5\frac{1}{2})\) en \(r = \sqrt{14\frac{1}{4}} \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 14 y + 24 = 0\)
\(x^{2} + (y + 7)^{2} - 49 + 24 = 0\)
\(x^{2} + (y + 7)^{2} = 25 \text{.}\)

Dus \(M (0 , -7)\) en \(r = \sqrt{25} = 5 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,-x - 2 y = c \\ A (-2 , 3)\end{rcases} c = -1 ⋅ -2 - 2 ⋅ 3 = -4\)
Dus \(n{:}\,-x - 2 y = -4 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}2 x - y = 3 \\ -x - 2 y = -4\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}2 x - y = 3 \\ -2 x - 4 y = -8\end{cases}\)
Optellen geeft \(-5 y = -5\) dus \(y = 1 \text{.}\)

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

\(d(A , l) = d(A , S) = \sqrt{(-2 - 2)^{2} + (3 - 1)^{2}} = \sqrt{20} \text{.}\)

\(A (-2 , 3)\) en \(r = d(A , l) = \sqrt{20} \text{,}\) dus
\(c{:}\,(x + 2)^{2} + (y - 3)^{2} = 20 \text{.}\)

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

Haakjes wegwerken geeft
\(x^{2} + y^{2} - 2 y + 1 = 25\)
en dus
\(c{:}\,x^{2} + y^{2} - 2 y - 24 = 0 \text{.}\)

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

\(\begin{rcases}y = 5 x + 2 \\ y_{M} = 3\end{rcases} \text{ geeft } 5 x + 2 = 3 \text{ dus } x_{M} = \frac{1}{5}\)

Middelpunt \(M_{1} (\frac{1}{5} , 3)\) en straal \(r = 3 \text{,}\) dus
\(c_{1}{:}\,(x - \frac{1}{5})^{2} + (y - 3)^{2} = 9\)

Op dezelfde manier geldt dat
\(\begin{rcases}y = 5 x + 2 \\ y_{M} = -3\end{rcases} \text{ geeft } 5 x + 2 = -3 \text{ dus } x_{M} = -1\)
Middelpunt \(M_{2} (-1 , -3)\) en straal \(r = 3 \text{,}\) dus
\(c_{2}{:}\,(x + 1)^{2} + (y + 3)^{2} = 9\)

\(x = 0\) invullen in de vergelijking van \(c\) geeft
\(0^{2} + y^{2} + 6 ⋅ 0 + 4 y + 3 = 0 \text{.}\)

De vergelijking oplossen geeft
\(y^{2} + 4 y + 3 = 0\)
\((y + 3) (y + 1) = 0\)
\(y = -3 ∨ y = -1\)
\(y_{A} > y_{B} \text{,}\) dus \(A (0 , -1)\) en \(B (0 , -3) \text{.}\)