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

\(r = d(M , A) = \sqrt{(-3 - -5)^{2} + (-2 - 1)^{2}} = \sqrt{13} \text{.}\)

\(c{:}\,(x + 3)^{2} + (y + 2)^{2} = 13 \text{.}\)

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

\(r = d(M , A) = \sqrt{(-\frac{1}{2} - 1)^{2} + (-3\frac{1}{2} - 0)^{2}} = \sqrt{14\frac{1}{2}} \text{.}\)

\(c{:}\,(x + \frac{1}{2})^{2} + (y + 3\frac{1}{2})^{2} = 14\frac{1}{2} \text{.}\)

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

\(c{:}\,(x - 7)^{2} + (y + 6)^{2} = 49 \text{.}\)

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

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

Dus \(M (4 , 2)\) en \(r = \sqrt{36} = 6 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 8 x + 11 y + 34 = 0\)
\((x + 4)^{2} - 16 + (y + 5\frac{1}{2})^{2} - 30\frac{1}{4} + 34 = 0\)
\((x + 4)^{2} + (y + 5\frac{1}{2})^{2} = 12\frac{1}{4} \text{.}\)

Dus \(M (-4 , -5\frac{1}{2})\) en \(r = \sqrt{12\frac{1}{4}} \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 8 y + 7 = 0\)
\(x^{2} + (y - 4)^{2} - 16 + 7 = 0\)
\(x^{2} + (y - 4)^{2} = 9 \text{.}\)

Dus \(M (0 , 4)\) en \(r = \sqrt{9} = 3 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,-5 x - 2 y = c \\ A (-5 , 4)\end{rcases} c = -5 ⋅ -5 - 2 ⋅ 4 = 17\)
Dus \(n{:}\,-5 x - 2 y = 17 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}2 x - 5 y = -1 \\ -5 x - 2 y = 17\end{cases}\) \(\begin{vmatrix}5 \\ 2\end{vmatrix}\) geeft \(\begin{cases}10 x - 25 y = -5 \\ -10 x - 4 y = 34\end{cases}\)
Optellen geeft \(-29 y = 29\) dus \(y = -1 \text{.}\)

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

\(d(A , l) = d(A , S) = \sqrt{(-5 - -3)^{2} + (4 - -1)^{2}} = \sqrt{29} \text{.}\)

\(A (-5 , 4)\) en \(r = d(A , l) = \sqrt{29} \text{,}\) dus
\(c{:}\,(x + 5)^{2} + (y - 4)^{2} = 29 \text{.}\)

\(c{:}\,x^{2} + (y + 5)^{2} = 16 \text{.}\)

Haakjes wegwerken geeft
\(x^{2} + y^{2} + 10 y + 25 = 16\)
en dus
\(c{:}\,x^{2} + y^{2} + 10 y + 9 = 0 \text{.}\)

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

\(\begin{rcases}y = 5 x + 3 \\ x_{M} = 2\end{rcases} \text{ geeft } y_{M} = 5 ⋅ 2 + 3 = 13\)

Middelpunt \(M_{1} (2 , 13)\) en straal \(r = 2 \text{,}\) dus
\(c_{1}{:}\,(x - 2)^{2} + (y - 13)^{2} = 4\)

Op dezelfde manier geldt dat
\(\begin{rcases}y = 5 x + 3 \\ x_{M} = -2\end{rcases} \text{ geeft } y_{M} = 5 ⋅ -2 + 3 = -7\)
Middelpunt \(M_{2} (-2 , -7)\) en straal \(r = 2 \text{,}\) dus
\(c_{2}{:}\,(x + 2)^{2} + (y + 7)^{2} = 4\)

\(x = 0\) invullen in de vergelijking van \(c\) geeft
\(0^{2} + y^{2} + 8 ⋅ 0 - 4 y + 3 = 0 \text{.}\)

De vergelijking oplossen geeft
\(y^{2} - 4 y + 3 = 0\)
\((y - 1) (y - 3) = 0\)
\(y = 1 ∨ y = 3\)
\(y_{A} > y_{B} \text{,}\) dus \(A (0 , 3)\) en \(B (0 , 1) \text{.}\)

Omschrijven van \(l\) geeft \(y = 2 x + 3 \text{.}\)
Substitutie in \(x^{2} + y^{2} + 2 x + 8 y + 7 = 0\) geeft
\(x^{2} + (2 x + 3)^{2} + 2 x + 8 (2 x + 3) + 7 = 0 \text{.}\)

Haakjes wegwerken geeft
\(x^{2} + 4 x^{2} + 12 x + 9 + 2 x + 16 x + 24 + 7 = 0\)
\(5 x^{2} + 30 x + 40 = 0 \text{.}\)

Oplossen van de vergelijking geeft
\(x^{2} + 6 x + 8 = 0\)
\((x + 4) (x + 2) = 0\)
Dus \(x = -4 ∨ x = -2 \text{.}\)

Invullen van \(x = -4\) in \(y = 2 x + 3\) geeft \(y = -5 \text{,}\) dus snijpunt \((-4 , -5) \text{.}\)
Invullen van \(x = -2\) geeft \(y = -1 \text{,}\) dus snijpunt \((-2 , -1) \text{.}\)