Kwadraatafsplitsen geeft \((x + 3)^{2} + (y - 4)^{2} = 20\)
Dus \(M (-3 , 4)\) en \(r = \sqrt{20} \text{.}\)

De lijn \(m\) door \(M\) en \(A\) heeft \(\text{rc}_{m} = {\Delta y \over \Delta x} = {4 - 6 \over -3 - 1} = \frac{1}{2} \text{.}\)

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

\(\begin{rcases}y = -2 x + b \\ \text{door } A (1 , 6)\end{rcases} \begin{matrix}6 = -2 ⋅ 1 + b \\ 6 = -2 + b \\ b = 8\end{matrix}\)
Dus \(l{:}\,y = -2 x + 8 \text{.}\)

\(\begin{rcases}c{:}\,x^{2} + y^{2} + 4 x + 6 y + 5 = 0 \\ x = 0\end{rcases}\) geeft
\(0^{2} + y^{2} + 4 ⋅ 0 + 6 y + 5 = 0\)
\(y^{2} + 6 y + 5 = 0\)
\((y + 5) (y + 1) = 0\)
\(y = -5 ∨ y = -1\)

\(y_{A} < y_{B} \text{,}\) dus \(A (0 , -5)\) en \(B (0 , -1) \text{.}\)

(kwadraatafsplitsen)
\((x + 2)^{2} - 4 + (y + 3)^{2} - 9 + 5 = 0\)
\((x + 2)^{2} + (y + 3)^{2} = 8\)
Dus \(M (-2 , -3) \text{.}\)

(voor de lijn door \(B\) en \(M\) geldt)
\(\text{rc}_{\text{BM}} = {\Delta y \over \Delta x} = {-3 - -1 \over -2 - 0} = 1 \text{.}\)

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

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

\(\begin{rcases}l{:}\,y = a x + b \\ A (14 , 9)\end{rcases} \begin{matrix}a ⋅ 14 + b = 9 \\ b = 9-14 a\end{matrix}\)
oftewel
\(l \text{: } y = a x+9-14 a\)
\(l \text{: } a x-y+9-14 a = 0\)

Van \(c\) is \(M (6 , 8)\) en \(r = 7 \text{.}\)
\(d(M , l) = \frac{\vert 6 a-8+9-14 a \vert}{\sqrt{a^{2}+(-1)^{2}}} = \frac{\vert -8 a+1 \vert}{\sqrt{a^{2}+1}} = 7\)

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

\(15 a^{2} - 16 a - 48 = 0\)
\(D = (-16)^{2} - 4 ⋅ 15 ⋅ -48 = 3\,136\)
\(a = {16 - \sqrt{3\,136} \over 2 ⋅ 15} ∨ a = {16 + \sqrt{3\,136} \over 2 ⋅ 15}\)
\(a = -1\frac{1}{3} ∨ a = 2\frac{2}{5}\)

\(a = -1\frac{1}{3}\) geeft \(b = 9-14 \cdot -1\frac{1}{3} = 27\frac{2}{3} \text{,}\) dus \(l_{1} \text{: } y = -1\frac{1}{3} x+27\frac{2}{3}\)
\(a = 2\frac{2}{5}\) geeft \(b = 9-14 \cdot 2\frac{2}{5} = -24\frac{3}{5} \text{,}\) dus \(l_{2} \text{: } y = 2\frac{2}{5} x-24\frac{3}{5}\)

\(l \text{: } y = -{1 \over 4} x+b\) geeft \(l \text{: } -{1 \over 4} x-y+b = 0\)

(kwadraatafsplitsen)
\((x-4)^{2}-16+y^{2}-137 = 0\)
\((x - 4)^{2} + y^{2} = 153\)
Dus \(M (4 , 0)\) en \(r = \sqrt{153} \text{.}\)

\(d(M , l) = \frac{\vert -{1 \over 4} \cdot 4-1 \cdot 0+b \vert}{\sqrt{(-{1 \over 4})^{2}+(-1)^{2}}} = \frac{\vert -1+b \vert}{\sqrt{{17 \over 16}}} = \sqrt{153}\)

(oplossen)
\(\vert -1+b \vert = \sqrt{153} \cdot \sqrt{{17 \over 16}} = 12\frac{3}{4}\)
\(-1+b = 12\frac{3}{4} ∨ -1+b = -12\frac{3}{4}\)
\(b = 1+12\frac{3}{4} = 13\frac{3}{4} ∨ b = 1-12\frac{3}{4} = -11\frac{3}{4}\)

\(b = 13\frac{3}{4}\) geeft \(l_{1} \text{: } y = -{1 \over 4} x+13\frac{3}{4}\)
\(b = -11\frac{3}{4}\) geeft \(l_{2} \text{: } y = -{1 \over 4} x-11\frac{3}{4}\)