\(c{:}\,(x + 7)^{2} + (y - 2)^{2} = 36 \text{.}\)

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

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

Het middelpunt \(M\) is het midden van \(A\kern{-.8pt}B \text{,}\) dus
\(M ({1 \over 2} (2 + -1) , {1 \over 2} (4 + -3)) = M (\frac{1}{2} , \frac{1}{2}) \text{.}\)

\(r = d(M , A) = \sqrt{(\frac{1}{2} - 2)^{2} + (\frac{1}{2} - 4)^{2}} = \sqrt{14\frac{1}{2}} \text{.}\)

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

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

\(c{:}\,(x + 4)^{2} + (y - 7)^{2} = 49 \text{.}\)

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

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 4 x - 10 y + 13 = 0\)
\((x - 2)^{2} - 4 + (y - 5)^{2} - 25 + 13 = 0\)
\((x - 2)^{2} + (y - 5)^{2} = 16 \text{.}\)

Dus \(M (2 , 5)\) en \(r = \sqrt{16} = 4 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 2 x + 3 y - 10 = 0\)
\((x - 1)^{2} - 1 + (y + 1\frac{1}{2})^{2} - 2\frac{1}{4} - 10 = 0\)
\((x - 1)^{2} + (y + 1\frac{1}{2})^{2} = 13\frac{1}{4} \text{.}\)

Dus \(M (1 , -1\frac{1}{2})\) en \(r = \sqrt{13\frac{1}{4}} \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 2 y - 8 = 0\)
\(x^{2} + (y + 1)^{2} - 1 - 8 = 0\)
\(x^{2} + (y + 1)^{2} = 9 \text{.}\)

Dus \(M (0 , -1)\) en \(r = \sqrt{9} = 3 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,-3 x - 2 y = c \\ A (2 , 4)\end{rcases} c = -3 ⋅ 2 - 2 ⋅ 4 = -14\)
Dus \(n{:}\,-3 x - 2 y = -14 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}2 x - 3 y = 5 \\ -3 x - 2 y = -14\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}6 x - 9 y = 15 \\ -6 x - 4 y = -28\end{cases}\)
Optellen geeft \(-13 y = -13\) dus \(y = 1 \text{.}\)

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

\(d(A , l) = d(A , S) = \sqrt{(2 - 4)^{2} + (4 - 1)^{2}} = \sqrt{13} \text{.}\)

\(A (2 , 4)\) en \(r = d(A , l) = \sqrt{13} \text{,}\) dus
\(c{:}\,(x - 2)^{2} + (y - 4)^{2} = 13 \text{.}\)

\(c{:}\,(x - 4)^{2} + (y - 1)^{2} = 49 \text{.}\)

Haakjes wegwerken geeft
\(x^{2} - 8 x + 16 + y^{2} - 2 y + 1 = 49\)
en dus
\(c{:}\,x^{2} + y^{2} - 8 x - 2 y - 32 = 0 \text{.}\)

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

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

Middelpunt \(M_{1} (4 , 13)\) en straal \(r = 4 \text{,}\) dus
\(c_{1}{:}\,(x - 4)^{2} + (y - 13)^{2} = 16\)

Op dezelfde manier geldt dat
\(\begin{rcases}y = 3 x + 1 \\ x_{M} = -4\end{rcases} \text{ geeft } y_{M} = 3 ⋅ -4 + 1 = -11\)
Middelpunt \(M_{2} (-4 , -11)\) en straal \(r = 4 \text{,}\) dus
\(c_{2}{:}\,(x + 4)^{2} + (y + 11)^{2} = 16\)

\(x = 2\) invullen in de vergelijking van \(c\) geeft
\(2^{2} + y^{2} + 4 ⋅ 2 - 2 y - 15 = 0 \text{.}\)

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

Omschrijven van \(l\) geeft \(y = -4 x - 4 \text{.}\)
Substitutie in \(x^{2} + y^{2} - 8 x + 6 y - 9 = 0\) geeft
\(x^{2} + (-4 x - 4)^{2} - 8 x + 6 (-4 x - 4) - 9 = 0 \text{.}\)

Haakjes wegwerken geeft
\(x^{2} + 16 x^{2} + 32 x + 16 - 8 x - 24 x - 24 - 9 = 0\)
\(17 x^{2} - 17 = 0 \text{.}\)

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

Invullen van \(x = 1\) in \(y = -4 x - 4\) geeft \(y = -8 \text{,}\) dus snijpunt \((1 , -8) \text{.}\)
Invullen van \(x = -1\) geeft \(y = 0 \text{,}\) dus snijpunt \((-1 , 0) \text{.}\)