\(c{:}\,(x + 1)^{2} + (y - 3)^{2} = 36 \text{.}\)

\(r = d(M , A) = \sqrt{(-2 - -7)^{2} + (4 - 7)^{2}} = \sqrt{34} \text{.}\)

\(c{:}\,(x + 2)^{2} + (y - 4)^{2} = 34 \text{.}\)

Het middelpunt \(M\) is het midden van \(A\kern{-.8pt}B \text{,}\) dus
\(M ({1 \over 2} (4 + -6) , {1 \over 2} (0 + -3)) = M (-1 , -1\frac{1}{2}) \text{.}\)

\(r = d(M , A) = \sqrt{(-1 - 4)^{2} + (-1\frac{1}{2} - 0)^{2}} = \sqrt{27\frac{1}{4}} \text{.}\)

\(c{:}\,(x + 1)^{2} + (y + 1\frac{1}{2})^{2} = 27\frac{1}{4} \text{.}\)

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

\(c{:}\,(x + 6)^{2} + (y + 3)^{2} = 9 \text{.}\)

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

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 10 x - 12 y + 57 = 0\)
\((x + 5)^{2} - 25 + (y - 6)^{2} - 36 + 57 = 0\)
\((x + 5)^{2} + (y - 6)^{2} = 4 \text{.}\)

Dus \(M (-5 , 6)\) en \(r = \sqrt{4} = 2 \text{.}\)

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

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

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 14 y + 33 = 0\)
\(x^{2} + (y - 7)^{2} - 49 + 33 = 0\)
\(x^{2} + (y - 7)^{2} = 16 \text{.}\)

Dus \(M (0 , 7)\) en \(r = \sqrt{16} = 4 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,2 x - y = c \\ A (-4 , -5)\end{rcases} c = 2 ⋅ -4 - 1 ⋅ -5 = -3\)
Dus \(n{:}\,2 x - y = -3 \text{.}\)

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

\(\begin{rcases}x + 2 y = -4 \\ y = -1\end{rcases} \begin{matrix}x + 2 ⋅ -1 = -4 \\ x = -2\end{matrix}\)
Dus \(S (-2 , -1) \text{.}\)

\(d(A , l) = d(A , S) = \sqrt{(-4 - -2)^{2} + (-5 - -1)^{2}} = \sqrt{20} \text{.}\)

\(A (-4 , -5)\) en \(r = d(A , l) = \sqrt{20} \text{,}\) dus
\(c{:}\,(x + 4)^{2} + (y + 5)^{2} = 20 \text{.}\)

\(c{:}\,(x + 3)^{2} + y^{2} = 49 \text{.}\)

Haakjes wegwerken geeft
\(x^{2} + 6 x + 9 + y^{2} = 49\)
en dus
\(c{:}\,x^{2} + y^{2} + 6 x - 40 = 0 \text{.}\)

De cirkels raken de \(y \text{-}\)as en hebben straal \(2 \text{,}\) dus \(x_{M} = 2\) of \(x_{M} = -2 \text{.}\)

\(\begin{rcases}y = 5 x + 3 \\ x_{M} = 2\end{rcases} \text{ geeft } y_{M} = 5 ⋅ 2 + 3 = 13\)

Middelpunt \(M_{1} (2 , 13)\) en straal \(r = 2 \text{,}\) dus
\(c_{1}{:}\,(x - 2)^{2} + (y - 13)^{2} = 4\)

Op dezelfde manier geldt dat
\(\begin{rcases}y = 5 x + 3 \\ x_{M} = -2\end{rcases} \text{ geeft } y_{M} = 5 ⋅ -2 + 3 = -7\)
Middelpunt \(M_{2} (-2 , -7)\) en straal \(r = 2 \text{,}\) dus
\(c_{2}{:}\,(x + 2)^{2} + (y + 7)^{2} = 4\)

\(x = 0\) invullen in de vergelijking van \(c\) geeft
\(0^{2} + y^{2} + 8 ⋅ 0 - 10 y + 16 = 0 \text{.}\)

De vergelijking oplossen geeft
\(y^{2} - 10 y + 16 = 0\)
\((y - 2) (y - 8) = 0\)
\(y = 2 ∨ y = 8\)
\(y_{A} > y_{B} \text{,}\) dus \(A (0 , 8)\) en \(B (0 , 2) \text{.}\)