Kwadraatafsplitsen geeft \((x - 4)^{2} + (y - 2)^{2} = 13\)
Dus \(M (4 , 2)\) en \(r = \sqrt{13} \text{.}\)

De lijn \(m\) door \(M\) en \(A\) heeft \(\text{rc}_{m} = {\Delta y \over \Delta x} = {2 - 5 \over 4 - 6} = 1\frac{1}{2} \text{.}\)

\(\begin{rcases}l \perp m \text{, dus } \text{rc}_{l} ⋅ \text{rc}_{m} = -1 \\ \text{rc}_{m} = 1\frac{1}{2}\end{rcases} \text{rc}_{l} = -\frac{2}{3}\)

\(\begin{rcases}y = -\frac{2}{3} x + b \\ \text{door } A (6 , 5)\end{rcases} \begin{matrix}5 = -\frac{2}{3} ⋅ 6 + b \\ 5 = -4 + b \\ b = 9\end{matrix}\)
Dus \(l{:}\,y = -\frac{2}{3} x + 9 \text{.}\)

\(\begin{rcases}c{:}\,x^{2} + y^{2} + 4 x - 16 = 0 \\ x = 0\end{rcases}\) geeft
\(0^{2} + y^{2} + 4 ⋅ 0 - 16 = 0\)
\(y^{2} - 16 = 0\)
\(y^{2} = 16\)
\(y = -4 ∨ y = 4\)

\(y_{A} < y_{B} \text{,}\) dus \(A (0 , -4)\) en \(B (0 , 4) \text{.}\)

(kwadraatafsplitsen)
\((x + 2)^{2} - 4 + y^{2} - 16 = 0\)
\((x + 2)^{2} + y^{2} = 20\)
Dus \(M (-2 , 0) \text{.}\)

(voor de lijn door \(B\) en \(M\) geldt)
\(\text{rc}_{\text{BM}} = {\Delta y \over \Delta x} = {0 - 4 \over -2 - 0} = 2 \text{.}\)

\(\begin{rcases}\text{BM} \perp l \text{, dus } \text{rc}_{\text{BM}} ⋅ \text{rc}_{l} = -1 \\ \text{rc}_{\text{BM}} = 2\end{rcases} \text{rc}_{l} = -\frac{1}{2}\)

\(\begin{rcases}l{:}\,y = -\frac{1}{2} x + b \\ \text{door } B (0 , 4)\end{rcases} \begin{matrix}4 = -\frac{1}{2} ⋅ 0 + b \\ 4 = 0 + b \\ b = 4\end{matrix}\)
Dus \(l{:}\,y = -\frac{1}{2} x + 4 \text{.}\)

\(\begin{rcases}l{:}\,y = a x + b \\ A (4 , 5)\end{rcases} \begin{matrix}a ⋅ 4 + b = 5 \\ b = 5-4 a\end{matrix}\)
oftewel
\(l \text{: } y = a x+5-4 a\)
\(l \text{: } a x-y+5-4 a = 0\)

Van \(c\) is \(M (-3 , 9)\) en \(r = 8 \text{.}\)
\(d(M , l) = \frac{\vert -3 a-9+5-4 a \vert}{\sqrt{a^{2}+(-1)^{2}}} = \frac{\vert -7 a-4 \vert}{\sqrt{a^{2}+1}} = 8\)

(oplossen)
\(\vert -7 a-4 \vert = 8 \sqrt{a^{2}+1}\)
\((\vert -7 a-4 \vert)^{2} = (8 \sqrt{a^{2}+1})^{2}\)
\(49 a^{2}+56 a+16 = 64 a^{2}+64\)

\(-15 a^{2} + 56 a - 48 = 0\)
\(15 a^{2} - 56 a + 48 = 0\)
\(D = (-56)^{2} - 4 ⋅ 15 ⋅ 48 = 256\)
\(a = {56 - \sqrt{256} \over 2 ⋅ 15} ∨ a = {56 + \sqrt{256} \over 2 ⋅ 15}\)
\(a = 1\frac{1}{3} ∨ a = 2\frac{2}{5}\)

\(a = 1\frac{1}{3}\) geeft \(b = 5-4 \cdot 1\frac{1}{3} = -\frac{1}{3} \text{,}\) dus \(l_{1} \text{: } y = 1\frac{1}{3} x-{1 \over 3}\)
\(a = 2\frac{2}{5}\) geeft \(b = 5-4 \cdot 2\frac{2}{5} = -4\frac{3}{5} \text{,}\) dus \(l_{2} \text{: } y = 2\frac{2}{5} x-4\frac{3}{5}\)

\(l \text{: } y = -{3 \over 4} x+b\) geeft \(l \text{: } -{3 \over 4} x-y+b = 0\)

(kwadraatafsplitsen)
\((x-2)^{2}-4+y^{2}-96 = 0\)
\((x - 2)^{2} + y^{2} = 100\)
Dus \(M (2 , 0)\) en \(r = 10 \text{.}\)

\(d(M , l) = \frac{\vert -{3 \over 4} \cdot 2-1 \cdot 0+b \vert}{\sqrt{(-{3 \over 4})^{2}+(-1)^{2}}} = \frac{\vert -{3 \over 2}+b \vert}{\sqrt{{25 \over 16}}} = 10\)

(oplossen)
\(\vert -{3 \over 2}+b \vert = 10 \sqrt{{25 \over 16}} = 12\frac{1}{2}\)
\(-{3 \over 2}+b = 12\frac{1}{2} ∨ -{3 \over 2}+b = -12\frac{1}{2}\)
\(b = {3 \over 2}+12\frac{1}{2} = 14 ∨ b = {3 \over 2}-12\frac{1}{2} = -11\)

\(b = 14\) geeft \(l_{1} \text{: } y = -{3 \over 4} x+14\)
\(b = -11\) geeft \(l_{2} \text{: } y = -{3 \over 4} x-11\)