\(c{:}\,(x + 3)^{2} + (y - 4)^{2} = 36 \text{.}\)

\(r = d(M , A) = \sqrt{(2 - 7)^{2} + (1 - 5)^{2}} = \sqrt{41} \text{.}\)

\(c{:}\,(x - 2)^{2} + (y - 1)^{2} = 41 \text{.}\)

Het middelpunt \(M\) is het midden van \(A\kern{-.8pt}B \text{,}\) dus
\(M ({1 \over 2} (-2 + 7) , {1 \over 2} (6 + -1)) = M (2\frac{1}{2} , 2\frac{1}{2}) \text{.}\)

\(r = d(M , A) = \sqrt{(2\frac{1}{2} - -2)^{2} + (2\frac{1}{2} - 6)^{2}} = \sqrt{32\frac{1}{2}} \text{.}\)

\(c{:}\,(x - 2\frac{1}{2})^{2} + (y - 2\frac{1}{2})^{2} = 32\frac{1}{2} \text{.}\)

Cirkel \(c\) raakt aan de \(x \text{-}\)as, dus \(r = d(M , x \text{-as}) = 7 \text{.}\)

\(c{:}\,(x + 3)^{2} + (y - 7)^{2} = 49 \text{.}\)

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

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 8 x - 4 y - 29 = 0\)
\((x + 4)^{2} - 16 + (y - 2)^{2} - 4 - 29 = 0\)
\((x + 4)^{2} + (y - 2)^{2} = 49 \text{.}\)

Dus \(M (-4 , 2)\) en \(r = \sqrt{49} = 7 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 6 x - 11 y + 34 = 0\)
\((x + 3)^{2} - 9 + (y - 5\frac{1}{2})^{2} - 30\frac{1}{4} + 34 = 0\)
\((x + 3)^{2} + (y - 5\frac{1}{2})^{2} = 5\frac{1}{4} \text{.}\)

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

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 6 y - 16 = 0\)
\(x^{2} + (y - 3)^{2} - 9 - 16 = 0\)
\(x^{2} + (y - 3)^{2} = 25 \text{.}\)

Dus \(M (0 , 3)\) en \(r = \sqrt{25} = 5 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,3 x - 4 y = c \\ A (-3 , -5)\end{rcases} c = 3 ⋅ -3 - 4 ⋅ -5 = 11\)
Dus \(n{:}\,3 x - 4 y = 11 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}4 x + 3 y = -2 \\ 3 x - 4 y = 11\end{cases}\) \(\begin{vmatrix}3 \\ 4\end{vmatrix}\) geeft \(\begin{cases}12 x + 9 y = -6 \\ 12 x - 16 y = 44\end{cases}\)
Aftrekken geeft \(25 y = -50\) dus \(y = -2 \text{.}\)

\(\begin{rcases}4 x + 3 y = -2 \\ y = -2\end{rcases} \begin{matrix}4 x + 3 ⋅ -2 = -2 \\ x = 1\end{matrix}\)
Dus \(S (1 , -2) \text{.}\)

\(d(A , l) = d(A , S) = \sqrt{(-3 - 1)^{2} + (-5 - -2)^{2}} = \sqrt{25} = 5 \text{.}\)

\(A (-3 , -5)\) en \(r = d(A , l) = \sqrt{25} \text{,}\) dus
\(c{:}\,(x + 3)^{2} + (y + 5)^{2} = 25 \text{.}\)

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

Haakjes wegwerken geeft
\(x^{2} + y^{2} - 10 y + 25 = 36\)
en dus
\(c{:}\,x^{2} + y^{2} - 10 y - 11 = 0 \text{.}\)

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

\(\begin{rcases}y = 5 x + 1 \\ x_{M} = 4\end{rcases} \text{ geeft } y_{M} = 5 ⋅ 4 + 1 = 21\)

Middelpunt \(M_{1} (4 , 21)\) en straal \(r = 4 \text{,}\) dus
\(c_{1}{:}\,(x - 4)^{2} + (y - 21)^{2} = 16\)

Op dezelfde manier geldt dat
\(\begin{rcases}y = 5 x + 1 \\ x_{M} = -4\end{rcases} \text{ geeft } y_{M} = 5 ⋅ -4 + 1 = -19\)
Middelpunt \(M_{2} (-4 , -19)\) en straal \(r = 4 \text{,}\) dus
\(c_{2}{:}\,(x + 4)^{2} + (y + 19)^{2} = 16\)

\(x = 4\) invullen in de vergelijking van \(c\) geeft
\(4^{2} + y^{2} - 4 ⋅ 4 + 6 y + 8 = 0 \text{.}\)

De vergelijking oplossen geeft
\(y^{2} + 6 y + 8 = 0\)
\((y + 4) (y + 2) = 0\)
\(y = -4 ∨ y = -2\)
\(y_{A} > y_{B} \text{,}\) dus \(A (4 , -2)\) en \(B (4 , -4) \text{.}\)