\(c{:}\,(x - 4)^{2} + (y - 5)^{2} = 49 \text{.}\)

\(r = d(M , A) = \sqrt{(-2 - -1)^{2} + (4 - 0)^{2}} = \sqrt{17} \text{.}\)

\(c{:}\,(x + 2)^{2} + (y - 4)^{2} = 17 \text{.}\)

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

\(r = d(M , A) = \sqrt{(-5\frac{1}{2} - -6)^{2} + (\frac{1}{2} - 0)^{2}} = \sqrt{\frac{1}{2}} \text{.}\)

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

Cirkel \(c\) raakt aan de \(x \text{-}\)as, dus \(r = d(M , x \text{-as}) = 2 \text{.}\)

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

\(c{:}\,x^{2} + (y - 4)^{2} = 25 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 8 x + 2 y - 8 = 0\)
\((x - 4)^{2} - 16 + (y + 1)^{2} - 1 - 8 = 0\)
\((x - 4)^{2} + (y + 1)^{2} = 25 \text{.}\)

Dus \(M (4 , -1)\) en \(r = \sqrt{25} = 5 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 7 x + 6 y + 16 = 0\)
\((x - 3\frac{1}{2})^{2} - 12\frac{1}{4} + (y + 3)^{2} - 9 + 16 = 0\)
\((x - 3\frac{1}{2})^{2} + (y + 3)^{2} = 5\frac{1}{4} \text{.}\)

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

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 2 x - 35 = 0\)
\((x + 1)^{2} - 1 + y^{2} - 35 = 0\)
\((x + 1)^{2} + y^{2} = 36 \text{.}\)

Dus \(M (-1 , 0)\) en \(r = \sqrt{36} = 6 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,-4 x - 2 y = c \\ A (-5 , 4)\end{rcases} c = -4 ⋅ -5 - 2 ⋅ 4 = 12\)
Dus \(n{:}\,-4 x - 2 y = 12 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}2 x - 4 y = 4 \\ -4 x - 2 y = 12\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}4 x - 8 y = 8 \\ -4 x - 2 y = 12\end{cases}\)
Optellen geeft \(-10 y = 20\) dus \(y = -2 \text{.}\)

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

\(d(A , l) = d(A , S) = \sqrt{(-5 - -2)^{2} + (4 - -2)^{2}} = \sqrt{45} \text{.}\)

\(A (-5 , 4)\) en \(r = d(A , l) = \sqrt{45} \text{,}\) dus
\(c{:}\,(x + 5)^{2} + (y - 4)^{2} = 45 \text{.}\)

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

Haakjes wegwerken geeft
\(x^{2} + 4 x + 4 + y^{2} + 2 y + 1 = 36\)
en dus
\(c{:}\,x^{2} + y^{2} + 4 x + 2 y - 31 = 0 \text{.}\)

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

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

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

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

\(x = 4\) invullen in de vergelijking van \(c\) geeft
\(4^{2} + y^{2} + 0 ⋅ 4 - 4 y - 21 = 0 \text{.}\)

De vergelijking oplossen geeft
\(y^{2} - 4 y - 5 = 0\)
\((y + 1) (y - 5) = 0\)
\(y = -1 ∨ y = 5\)
\(y_{A} > y_{B} \text{,}\) dus \(A (4 , 5)\) en \(B (4 , -1) \text{.}\)