\(c{:}\,(x + 2)^{2} + (y + 3)^{2} = 25 \text{.}\)

\(r = d(M , A) = \sqrt{(1 - 4)^{2} + (2 - 0)^{2}} = \sqrt{13} \text{.}\)

\(c{:}\,(x - 1)^{2} + (y - 2)^{2} = 13 \text{.}\)

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

\(r = d(M , A) = \sqrt{(3\frac{1}{2} - 7)^{2} + (-1\frac{1}{2} - -5)^{2}} = \sqrt{24\frac{1}{2}} \text{.}\)

\(c{:}\,(x - 3\frac{1}{2})^{2} + (y + 1\frac{1}{2})^{2} = 24\frac{1}{2} \text{.}\)

Cirkel \(c\) raakt aan de \(x \text{-}\)as, dus \(r = d(M , x \text{-as}) = 4 \text{.}\)

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

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

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} + 10 x + 12 y + 12 = 0\)
\((x + 5)^{2} - 25 + (y + 6)^{2} - 36 + 12 = 0\)
\((x + 5)^{2} + (y + 6)^{2} = 49 \text{.}\)

Dus \(M (-5 , -6)\) en \(r = \sqrt{49} = 7 \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 5 x - 8 y + 13 = 0\)
\((x - 2\frac{1}{2})^{2} - 6\frac{1}{4} + (y - 4)^{2} - 16 + 13 = 0\)
\((x - 2\frac{1}{2})^{2} + (y - 4)^{2} = 9\frac{1}{4} \text{.}\)

Dus \(M (2\frac{1}{2} , 4)\) en \(r = \sqrt{9\frac{1}{4}} \text{.}\)

Kwadraatafsplitsen geeft
\(x^{2} + y^{2} - 14 x + 13 = 0\)
\((x - 7)^{2} - 49 + y^{2} + 13 = 0\)
\((x - 7)^{2} + y^{2} = 36 \text{.}\)

Dus \(M (7 , 0)\) en \(r = \sqrt{36} = 6 \text{.}\)

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

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

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

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

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

\(c{:}\,(x + 1)^{2} + (y + 7)^{2} = 9 \text{.}\)

Haakjes wegwerken geeft
\(x^{2} + 2 x + 1 + y^{2} + 14 y + 49 = 9\)
en dus
\(c{:}\,x^{2} + y^{2} + 2 x + 14 y + 41 = 0 \text{.}\)

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

\(\begin{rcases}y = 2 x + 5 \\ y_{M} = 4\end{rcases} \text{ geeft } 2 x + 5 = 4 \text{ dus } x_{M} = -\frac{1}{2}\)

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

Op dezelfde manier geldt dat
\(\begin{rcases}y = 2 x + 5 \\ y_{M} = -4\end{rcases} \text{ geeft } 2 x + 5 = -4 \text{ dus } x_{M} = -4\frac{1}{2}\)
Middelpunt \(M_{2} (-4\frac{1}{2} , -4)\) en straal \(r = 4 \text{,}\) dus
\(c_{2}{:}\,(x + 4\frac{1}{2})^{2} + (y + 4)^{2} = 16\)

\(x = 3\) invullen in de vergelijking van \(c\) geeft
\(3^{2} + y^{2} + 0 ⋅ 3 - 2 y - 12 = 0 \text{.}\)

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 \(A (3 , 3)\) en \(B (3 , -1) \text{.}\)

Omschrijven van \(l\) geeft \(x = 3 y - 4 \text{.}\)
Substitutie in \(x^{2} + y^{2} + 10 x - 6 y + 14 = 0\) geeft
\((3 y - 4)^{2} + y^{2} + 10 (3 y - 4) - 6 y + 14 = 0 \text{.}\)

Haakjes wegwerken geeft
\(9 y^{2} - 24 y + 16 + y^{2} + 30 y - 40 - 6 y + 14 = 0\)
\(10 y^{2} - 10 = 0 \text{.}\)

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

Invullen van \(y = 1\) in \(x = 3 y - 4\) geeft \(x = -1 \text{,}\) dus snijpunt \((-1 , 1) \text{.}\)
Invullen van \(y = -1\) geeft \(x = -7 \text{,}\) dus snijpunt \((-7 , -1) \text{.}\)