\(c{:}\,(x + 5)^{2} + (y - 1)^{2} = 4 \text{.}\)

\(r = d(M , A) = \sqrt{(2 - -3)^{2} + (3 - 7)^{2}} = \sqrt{41} \text{.}\)

\(c{:}\,(x - 2)^{2} + (y - 3)^{2} = 41 \text{.}\)

Het middelpunt \(M\) is het midden van \(A\kern{-.8pt}B \text{,}\) dus
\(M ({1 \over 2} (-7 + -6) , {1 \over 2} (-4 + 5)) = M (-6\frac{1}{2} , \frac{1}{2}) \text{.}\)

\(r = d(M , A) = \sqrt{(-6\frac{1}{2} - -7)^{2} + (\frac{1}{2} - -4)^{2}} = \sqrt{20\frac{1}{2}} \text{.}\)

\(c{:}\,(x + 6\frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = 20\frac{1}{2} \text{.}\)

Cirkel \(c\) raakt aan de \(y \text{-}\)as, dus \(r = d(M , y \text{-as}) = 1 \text{.}\)

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

\(c{:}\,x^{2} + (y + 6)^{2} = 49 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 6 x + 12 y - 4 = 0\)
\((x - 3)^{2} - 9 + (y + 6)^{2} - 36 - 4 = 0\)
\((x - 3)^{2} + (y + 6)^{2} = 49 \text{.}\)

Dus \(M (3 , -6)\) en \(r = \sqrt{49} = 7 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 8 x - 3 y + 1 = 0\)
\((x - 4)^{2} - 16 + (y - 1\frac{1}{2})^{2} - 2\frac{1}{4} + 1 = 0\)
\((x - 4)^{2} + (y - 1\frac{1}{2})^{2} = 17\frac{1}{4} \text{.}\)

Dus \(M (4 , 1\frac{1}{2})\) en \(r = \sqrt{17\frac{1}{4}} \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 8 x + 12 = 0\)
\((x - 4)^{2} - 16 + y^{2} + 12 = 0\)
\((x - 4)^{2} + y^{2} = 4 \text{.}\)

Dus \(M (4 , 0)\) en \(r = \sqrt{4} = 2 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,x - 4 y = c \\ A (-2 , -4)\end{rcases} c = 1 ⋅ -2 - 4 ⋅ -4 = 14\)
Dus \(n{:}\,x - 4 y = 14 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}4 x + y = 5 \\ x - 4 y = 14\end{cases}\) \(\begin{vmatrix}1 \\ 4\end{vmatrix}\) geeft \(\begin{cases}4 x + y = 5 \\ 4 x - 16 y = 56\end{cases}\)
Aftrekken geeft \(17 y = -51\) dus \(y = -3 \text{.}\)

\(\begin{rcases}4 x + y = 5 \\ y = -3\end{rcases} \begin{matrix}4 x + 1 ⋅ -3 = 5 \\ x = 2\end{matrix}\)
Dus \(S (2 , -3) \text{.}\)

\(d(A , l) = d(A , S) = \sqrt{(-2 - 2)^{2} + (-4 - -3)^{2}} = \sqrt{17} \text{.}\)

\(A (-2 , -4)\) en \(r = d(A , l) = \sqrt{17} \text{,}\) dus
\(c{:}\,(x + 2)^{2} + (y + 4)^{2} = 17 \text{.}\)

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

Haakjes wegwerken geeft
\(x^{2} - 4 x + 4 + y^{2} = 25\)
en dus
\(c{:}\,x^{2} + y^{2} - 4 x - 21 = 0 \text{.}\)

De cirkels raken de \(y \text{-}\)as en hebben straal \(3 \text{,}\) dus \(x_{M} = 3\) of \(x_{M} = -3 \text{.}\)

\(\begin{rcases}y = 4 x + 1 \\ x_{M} = 3\end{rcases} \text{ geeft } y_{M} = 4 ⋅ 3 + 1 = 13\)

Middelpunt \(M_{1} (3 , 13)\) en straal \(r = 3 \text{,}\) dus
\(c_{1}{:}\,(x - 3)^{2} + (y - 13)^{2} = 9\)

Op dezelfde manier geldt dat
\(\begin{rcases}y = 4 x + 1 \\ x_{M} = -3\end{rcases} \text{ geeft } y_{M} = 4 ⋅ -3 + 1 = -11\)
Middelpunt \(M_{2} (-3 , -11)\) en straal \(r = 3 \text{,}\) dus
\(c_{2}{:}\,(x + 3)^{2} + (y + 11)^{2} = 9\)

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

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

Omschrijven van \(l\) geeft \(x = 2 y - 4 \text{.}\)
Substitutie in \(x^{2} + y^{2} + 8 x + 10 y + 1 = 0\) geeft
\((2 y - 4)^{2} + y^{2} + 8 (2 y - 4) + 10 y + 1 = 0 \text{.}\)

Haakjes wegwerken geeft
\(4 y^{2} - 16 y + 16 + y^{2} + 16 y - 32 + 10 y + 1 = 0\)
\(5 y^{2} + 10 y - 15 = 0 \text{.}\)

Oplossen van de vergelijking geeft
\(y^{2} + 2 y - 3 = 0\)
\((y + 3) (y - 1) = 0\)
Dus \(y = -3 ∨ y = 1 \text{.}\)

Invullen van \(y = -3\) in \(x = 2 y - 4\) geeft \(x = -10 \text{,}\) dus snijpunt \((-10 , -3) \text{.}\)
Invullen van \(y = 1\) geeft \(x = -2 \text{,}\) dus snijpunt \((-2 , 1) \text{.}\)