\(c{:}\,(x - 2)^{2} + (y + 6)^{2} = 25 \text{.}\)

\(r = d(M , A) = \sqrt{(1 - 5)^{2} + (-3 - -5)^{2}} = \sqrt{20} \text{.}\)

\(c{:}\,(x - 1)^{2} + (y + 3)^{2} = 20 \text{.}\)

Het middelpunt \(M\) is het midden van \(A\kern{-.8pt}B \text{,}\) dus
\(M ({1 \over 2} (0 + -3) , {1 \over 2} (-5 + 1)) = M (-1\frac{1}{2} , -2) \text{.}\)

\(r = d(M , A) = \sqrt{(-1\frac{1}{2} - 0)^{2} + (-2 - -5)^{2}} = \sqrt{11\frac{1}{4}} \text{.}\)

\(c{:}\,(x + 1\frac{1}{2})^{2} + (y + 2)^{2} = 11\frac{1}{4} \text{.}\)

Cirkel \(c\) raakt aan de \(y \text{-}\)as, dus \(r = d(M , y \text{-as}) = 5 \text{.}\)

\(c{:}\,(x - 5)^{2} + (y + 6)^{2} = 25 \text{.}\)

\(c{:}\,x^{2} + (y + 4)^{2} = 36 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 6 x + 4 y - 12 = 0\)
\((x + 3)^{2} - 9 + (y + 2)^{2} - 4 - 12 = 0\)
\((x + 3)^{2} + (y + 2)^{2} = 25 \text{.}\)

Dus \(M (-3 , -2)\) en \(r = \sqrt{25} = 5 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 12 x + 7 y + 43 = 0\)
\((x + 6)^{2} - 36 + (y + 3\frac{1}{2})^{2} - 12\frac{1}{4} + 43 = 0\)
\((x + 6)^{2} + (y + 3\frac{1}{2})^{2} = 5\frac{1}{4} \text{.}\)

Dus \(M (-6 , -3\frac{1}{2})\) en \(r = \sqrt{5\frac{1}{4}} \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 12 x + 27 = 0\)
\((x - 6)^{2} - 36 + y^{2} + 27 = 0\)
\((x - 6)^{2} + y^{2} = 9 \text{.}\)

Dus \(M (6 , 0)\) en \(r = \sqrt{9} = 3 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,-5 x - 3 y = c \\ A (4 , -4)\end{rcases} c = -5 ⋅ 4 - 3 ⋅ -4 = -8\)
Dus \(n{:}\,-5 x - 3 y = -8 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}3 x - 5 y = -2 \\ -5 x - 3 y = -8\end{cases}\) \(\begin{vmatrix}5 \\ 3\end{vmatrix}\) geeft \(\begin{cases}15 x - 25 y = -10 \\ -15 x - 9 y = -24\end{cases}\)
Optellen geeft \(-34 y = -34\) dus \(y = 1 \text{.}\)

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

\(d(A , l) = d(A , S) = \sqrt{(4 - 1)^{2} + (-4 - 1)^{2}} = \sqrt{34} \text{.}\)

\(A (4 , -4)\) en \(r = d(A , l) = \sqrt{34} \text{,}\) dus
\(c{:}\,(x - 4)^{2} + (y + 4)^{2} = 34 \text{.}\)

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

Haakjes wegwerken geeft
\(x^{2} + 4 x + 4 + y^{2} - 12 y + 36 = 9\)
en dus
\(c{:}\,x^{2} + y^{2} + 4 x - 12 y + 31 = 0 \text{.}\)

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

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

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

Op dezelfde manier geldt dat
\(\begin{rcases}y = 5 x + 3 \\ y_{M} = -4\end{rcases} \text{ geeft } 5 x + 3 = -4 \text{ dus } x_{M} = -1\frac{2}{5}\)
Middelpunt \(M_{2} (-1\frac{2}{5} , -4)\) en straal \(r = 4 \text{,}\) dus
\(c_{2}{:}\,(x + 1\frac{2}{5})^{2} + (y + 4)^{2} = 16\)

\(x = 7\) invullen in de vergelijking van \(c\) geeft
\(7^{2} + y^{2} - 10 ⋅ 7 + 8 y + 36 = 0 \text{.}\)

De vergelijking oplossen geeft
\(y^{2} + 8 y + 15 = 0\)
\((y + 5) (y + 3) = 0\)
\(y = -5 ∨ y = -3\)
\(y_{A} > y_{B} \text{,}\) dus \(A (7 , -3)\) en \(B (7 , -5) \text{.}\)