\(c{:}\,(x + 6)^{2} + (y + 5)^{2} = 4 \text{.}\)

\(r = d(M , A) = \sqrt{(-1 - -3)^{2} + (3 - 4)^{2}} = \sqrt{5} \text{.}\)

\(c{:}\,(x + 1)^{2} + (y - 3)^{2} = 5 \text{.}\)

Het middelpunt \(M\) is het midden van \(A\kern{-.8pt}B \text{,}\) dus
\(M ({1 \over 2} (-1 + -5) , {1 \over 2} (7 + 0)) = M (-3 , 3\frac{1}{2}) \text{.}\)

\(r = d(M , A) = \sqrt{(-3 - -1)^{2} + (3\frac{1}{2} - 7)^{2}} = \sqrt{16\frac{1}{4}} \text{.}\)

\(c{:}\,(x + 3)^{2} + (y - 3\frac{1}{2})^{2} = 16\frac{1}{4} \text{.}\)

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

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

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

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 12 x + 8 y + 27 = 0\)
\((x + 6)^{2} - 36 + (y + 4)^{2} - 16 + 27 = 0\)
\((x + 6)^{2} + (y + 4)^{2} = 25 \text{.}\)

Dus \(M (-6 , -4)\) en \(r = \sqrt{25} = 5 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 10 x + 3 y + 20 = 0\)
\((x + 5)^{2} - 25 + (y + 1\frac{1}{2})^{2} - 2\frac{1}{4} + 20 = 0\)
\((x + 5)^{2} + (y + 1\frac{1}{2})^{2} = 7\frac{1}{4} \text{.}\)

Dus \(M (-5 , -1\frac{1}{2})\) en \(r = \sqrt{7\frac{1}{4}} \text{.}\)

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

Dus \(M (-5 , 0)\) en \(r = \sqrt{9} = 3 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,-x - 3 y = c \\ A (4 , 5)\end{rcases} c = -1 ⋅ 4 - 3 ⋅ 5 = -19\)
Dus \(n{:}\,-x - 3 y = -19 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}3 x - y = -3 \\ -x - 3 y = -19\end{cases}\) \(\begin{vmatrix}1 \\ 3\end{vmatrix}\) geeft \(\begin{cases}3 x - y = -3 \\ -3 x - 9 y = -57\end{cases}\)
Optellen geeft \(-10 y = -60\) dus \(y = 6 \text{.}\)

\(\begin{rcases}3 x - y = -3 \\ y = 6\end{rcases} \begin{matrix}3 x - 1 ⋅ 6 = -3 \\ x = 1\end{matrix}\)
Dus \(S (1 , 6) \text{.}\)

\(d(A , l) = d(A , S) = \sqrt{(4 - 1)^{2} + (5 - 6)^{2}} = \sqrt{10} \text{.}\)

\(A (4 , 5)\) en \(r = d(A , l) = \sqrt{10} \text{,}\) dus
\(c{:}\,(x - 4)^{2} + (y - 5)^{2} = 10 \text{.}\)

\(c{:}\,(x - 1)^{2} + (y + 2)^{2} = 9 \text{.}\)

Haakjes wegwerken geeft
\(x^{2} - 2 x + 1 + y^{2} + 4 y + 4 = 9\)
en dus
\(c{:}\,x^{2} + y^{2} - 2 x + 4 y - 4 = 0 \text{.}\)

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

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

Middelpunt \(M_{1} (\frac{1}{5} , 3)\) en straal \(r = 3 \text{,}\) dus
\(c_{1}{:}\,(x - \frac{1}{5})^{2} + (y - 3)^{2} = 9\)

Op dezelfde manier geldt dat
\(\begin{rcases}y = 5 x + 2 \\ y_{M} = -3\end{rcases} \text{ geeft } 5 x + 2 = -3 \text{ dus } x_{M} = -1\)
Middelpunt \(M_{2} (-1 , -3)\) en straal \(r = 3 \text{,}\) dus
\(c_{2}{:}\,(x + 1)^{2} + (y + 3)^{2} = 9\)

\(x = 1\) invullen in de vergelijking van \(c\) geeft
\(1^{2} + y^{2} + 4 ⋅ 1 + 2 y - 20 = 0 \text{.}\)

De vergelijking oplossen geeft
\(y^{2} + 2 y - 15 = 0\)
\((y + 5) (y - 3) = 0\)
\(y = -5 ∨ y = 3\)
\(y_{A} > y_{B} \text{,}\) dus \(A (1 , 3)\) en \(B (1 , -5) \text{.}\)

Omschrijven van \(l\) geeft \(x = 4 y + 5 \text{.}\)
Substitutie in \(x^{2} + y^{2} - 4 x + 10 y - 5 = 0\) geeft
\((4 y + 5)^{2} + y^{2} - 4 (4 y + 5) + 10 y - 5 = 0 \text{.}\)

Haakjes wegwerken geeft
\(16 y^{2} + 40 y + 25 + y^{2} - 16 y - 20 + 10 y - 5 = 0\)
\(17 y^{2} + 34 y = 0 \text{.}\)

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

Invullen van \(y = -2\) in \(x = 4 y + 5\) geeft \(x = -3 \text{,}\) dus snijpunt \((-3 , -2) \text{.}\)
Invullen van \(y = 0\) geeft \(x = 5 \text{,}\) dus snijpunt \((5 , 0) \text{.}\)