\(c{:}\,(x - 2)^{2} + (y - 3)^{2} = 36 \text{.}\)

\(r = d(M , A) = \sqrt{(3 - 8)^{2} + (-2 - 2)^{2}} = \sqrt{41} \text{.}\)

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

Het middelpunt \(M\) is het midden van \(A\kern{-.8pt}B \text{,}\) dus
\(M ({1 \over 2} (7 + -3) , {1 \over 2} (-2 + -6)) = M (2 , -4) \text{.}\)

\(r = d(M , A) = \sqrt{(2 - 7)^{2} + (-4 - -2)^{2}} = \sqrt{29} \text{.}\)

\(c{:}\,(x - 2)^{2} + (y + 4)^{2} = 29 \text{.}\)

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

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

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

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 10 x + 6 y + 18 = 0\)
\((x + 5)^{2} - 25 + (y + 3)^{2} - 9 + 18 = 0\)
\((x + 5)^{2} + (y + 3)^{2} = 16 \text{.}\)

Dus \(M (-5 , -3)\) en \(r = \sqrt{16} = 4 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 11 x + 8 y + 41 = 0\)
\((x + 5\frac{1}{2})^{2} - 30\frac{1}{4} + (y + 4)^{2} - 16 + 41 = 0\)
\((x + 5\frac{1}{2})^{2} + (y + 4)^{2} = 5\frac{1}{4} \text{.}\)

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

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

Dus \(M (0 , -2)\) en \(r = \sqrt{16} = 4 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,-x - 2 y = c \\ A (3 , -5)\end{rcases} c = -1 ⋅ 3 - 2 ⋅ -5 = 7\)
Dus \(n{:}\,-x - 2 y = 7 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}2 x - y = 1 \\ -x - 2 y = 7\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}2 x - y = 1 \\ -2 x - 4 y = 14\end{cases}\)
Optellen geeft \(-5 y = 15\) dus \(y = -3 \text{.}\)

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

\(d(A , l) = d(A , S) = \sqrt{(3 - -1)^{2} + (-5 - -3)^{2}} = \sqrt{20} \text{.}\)

\(A (3 , -5)\) en \(r = d(A , l) = \sqrt{20} \text{,}\) dus
\(c{:}\,(x - 3)^{2} + (y + 5)^{2} = 20 \text{.}\)

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

Haakjes wegwerken geeft
\(x^{2} + 12 x + 36 + y^{2} = 4\)
en dus
\(c{:}\,x^{2} + y^{2} + 12 x + 32 = 0 \text{.}\)

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

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

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

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

\(x = 8\) invullen in de vergelijking van \(c\) geeft
\(8^{2} + y^{2} - 8 ⋅ 8 - 1 = 0 \text{.}\)

De vergelijking oplossen geeft
\(y^{2} - 1 = 0\)
\((y + 1) (y - 1) = 0\)
\(y = -1 ∨ y = 1\)
\(y_{A} > y_{B} \text{,}\) dus \(A (8 , 1)\) en \(B (8 , -1) \text{.}\)