Kwadraatafsplitsen geeft \((x + 3)^{2} + (y - 2)^{2} = 13\)
Dus \(M (-3 , 2)\) en \(r = \sqrt{13} \text{.}\)

De lijn \(m\) door \(M\) en \(A\) heeft \(\text{rc}_{m} = {\Delta y \over \Delta x} = {2 - 4 \over -3 - 0} = \frac{2}{3} \text{.}\)

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

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

\(\begin{rcases}c{:}\,x^{2} + y^{2} - 4 x - 8 y + 18 = 0 \\ x = 3\end{rcases}\) geeft
\(3^{2} + y^{2} - 4 ⋅ 3 - 8 y + 18 = 0\)
\(y^{2} - 8 y + 15 = 0\)
\((y - 3) (y - 5) = 0\)
\(y = 3 ∨ y = 5\)

\(y_{A} > y_{B} \text{,}\) dus \(A (3 , 5)\) en \(B (3 , 3) \text{.}\)

(kwadraatafsplitsen)
\((x - 2)^{2} - 4 + (y - 4)^{2} - 16 + 18 = 0\)
\((x - 2)^{2} + (y - 4)^{2} = 2\)
Dus \(M (2 , 4) \text{.}\)

(voor de lijn door \(B\) en \(M\) geldt)
\(\text{rc}_{\text{BM}} = {\Delta y \over \Delta x} = {4 - 3 \over 2 - 3} = -1 \text{.}\)

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

\(\begin{rcases}k{:}\,y = x + b \\ \text{door } B (3 , 3)\end{rcases} \begin{matrix}3 = 1 ⋅ 3 + b \\ 3 = 3 + b \\ b = 0\end{matrix}\)
Dus \(k{:}\,y = x \text{.}\)

\(\begin{rcases}l{:}\,y = a x + b \\ A (-1 , 4)\end{rcases} \begin{matrix}a ⋅ -1 + b = 4 \\ b = a+4\end{matrix}\)
oftewel
\(l \text{: } y = a x+a+4\)
\(l \text{: } a x-y+a+4 = 0\)

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

(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 \(b = -1\frac{1}{3}+4 = 2\frac{2}{3} \text{,}\) dus \(l_{1} \text{: } y = -1\frac{1}{3} x+2\frac{2}{3}\)
\(a = 2\frac{2}{5}\) geeft \(b = 2\frac{2}{5}+4 = 6\frac{2}{5} \text{,}\) dus \(l_{2} \text{: } y = 2\frac{2}{5} x+6\frac{2}{5}\)

\(l \text{: } y = {1 \over 3} x+b\) geeft \(l \text{: } {1 \over 3} x-y+b = 0\)

(kwadraatafsplitsen)
\((x+2)^{2}-4+(y-1)^{2}-1-35 = 0\)
\((x + 2)^{2} + (y - 1)^{2} = 40\)
Dus \(M (-2 , 1)\) en \(r = \sqrt{40} \text{.}\)

\(d(M , l) = \frac{\vert {1 \over 3} \cdot -2-1 \cdot 1+b \vert}{\sqrt{({1 \over 3})^{2}+(-1)^{2}}} = \frac{\vert -{5 \over 3}+b \vert}{\sqrt{{10 \over 9}}} = \sqrt{40}\)

(oplossen)
\(\vert -{5 \over 3}+b \vert = \sqrt{40} \cdot \sqrt{{10 \over 9}} = 6\frac{2}{3}\)
\(-{5 \over 3}+b = 6\frac{2}{3} ∨ -{5 \over 3}+b = -6\frac{2}{3}\)
\(b = {5 \over 3}+6\frac{2}{3} = 8\frac{1}{3} ∨ b = {5 \over 3}-6\frac{2}{3} = -5\)

\(b = 8\frac{1}{3}\) geeft \(l_{1} \text{: } y = {1 \over 3} x+8\frac{1}{3}\)
\(b = -5\) geeft \(l_{2} \text{: } y = {1 \over 3} x-5\)