\(c{:}\,(x - 2)^{2} + (y + 7)^{2} = 16 \text{.}\)

\(r = d(M , A) = \sqrt{(2 - 5)^{2} + (3 - 8)^{2}} = \sqrt{34} \text{.}\)

\(c{:}\,(x - 2)^{2} + (y - 3)^{2} = 34 \text{.}\)

Het middelpunt \(M\) is het midden van \(A\kern{-.8pt}B \text{,}\) dus
\(M ({1 \over 2} (0 + 4) , {1 \over 2} (5 + -7)) = M (2 , -1) \text{.}\)

\(r = d(M , A) = \sqrt{(2 - 0)^{2} + (-1 - 5)^{2}} = \sqrt{40} \text{.}\)

\(c{:}\,(x - 2)^{2} + (y + 1)^{2} = 40 \text{.}\)

Cirkel \(c\) raakt aan de \(x \text{-}\)as, dus \(r = d(M , x \text{-as}) = 6 \text{.}\)

\(c{:}\,(x + 1)^{2} + (y + 6)^{2} = 36 \text{.}\)

\(c{:}\,x^{2} + (y + 7)^{2} = 16 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 8 x - 14 y + 56 = 0\)
\((x + 4)^{2} - 16 + (y - 7)^{2} - 49 + 56 = 0\)
\((x + 4)^{2} + (y - 7)^{2} = 9 \text{.}\)

Dus \(M (-4 , 7)\) en \(r = \sqrt{9} = 3 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 11 x - 10 y + 39 = 0\)
\((x - 5\frac{1}{2})^{2} - 30\frac{1}{4} + (y - 5)^{2} - 25 + 39 = 0\)
\((x - 5\frac{1}{2})^{2} + (y - 5)^{2} = 16\frac{1}{4} \text{.}\)

Dus \(M (5\frac{1}{2} , 5)\) en \(r = \sqrt{16\frac{1}{4}} \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 10 y + 9 = 0\)
\(x^{2} + (y - 5)^{2} - 25 + 9 = 0\)
\(x^{2} + (y - 5)^{2} = 16 \text{.}\)

Dus \(M (0 , 5)\) en \(r = \sqrt{16} = 4 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,-3 x - 2 y = c \\ A (-4 , 2)\end{rcases} c = -3 ⋅ -4 - 2 ⋅ 2 = 8\)
Dus \(n{:}\,-3 x - 2 y = 8 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}2 x - 3 y = -1 \\ -3 x - 2 y = 8\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}6 x - 9 y = -3 \\ -6 x - 4 y = 16\end{cases}\)
Optellen geeft \(-13 y = 13\) dus \(y = -1 \text{.}\)

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

\(d(A , l) = d(A , S) = \sqrt{(-4 - -2)^{2} + (2 - -1)^{2}} = \sqrt{13} \text{.}\)

\(A (-4 , 2)\) en \(r = d(A , l) = \sqrt{13} \text{,}\) dus
\(c{:}\,(x + 4)^{2} + (y - 2)^{2} = 13 \text{.}\)

\(c{:}\,(x + 7)^{2} + (y - 5)^{2} = 9 \text{.}\)

Haakjes wegwerken geeft
\(x^{2} + 14 x + 49 + y^{2} - 10 y + 25 = 9\)
en dus
\(c{:}\,x^{2} + y^{2} + 14 x - 10 y + 65 = 0 \text{.}\)

De cirkels raken de \(x \text{-}\)as en hebben straal \(4 \text{,}\) dus \(y_{M} = 4\) of \(y_{M} = -4 \text{.}\)

\(\begin{rcases}y = 5 x + 1 \\ y_{M} = 4\end{rcases} \text{ geeft } 5 x + 1 = 4 \text{ dus } x_{M} = \frac{3}{5}\)

Middelpunt \(M_{1} (\frac{3}{5} , 4)\) en straal \(r = 4 \text{,}\) dus
\(c_{1}{:}\,(x - \frac{3}{5})^{2} + (y - 4)^{2} = 16\)

Op dezelfde manier geldt dat
\(\begin{rcases}y = 5 x + 1 \\ y_{M} = -4\end{rcases} \text{ geeft } 5 x + 1 = -4 \text{ dus } x_{M} = -1\)
Middelpunt \(M_{2} (-1 , -4)\) en straal \(r = 4 \text{,}\) dus
\(c_{2}{:}\,(x + 1)^{2} + (y + 4)^{2} = 16\)

\(x = 9\) invullen in de vergelijking van \(c\) geeft
\(9^{2} + y^{2} - 10 ⋅ 9 + 2 y + 1 = 0 \text{.}\)

De vergelijking oplossen geeft
\(y^{2} + 2 y - 8 = 0\)
\((y + 4) (y - 2) = 0\)
\(y = -4 ∨ y = 2\)
\(y_{A} > y_{B} \text{,}\) dus \(A (9 , 2)\) en \(B (9 , -4) \text{.}\)