Raaklijnen aan cirkels

Kwadraatafsplitsen geeft \((x - 1)^{2} + (y - 3)^{2} = 25\)
Dus \(M (1 , 3)\) en \(r = \sqrt{25} = 5 \text{.}\)

De lijn \(m\) door \(M\) en \(A\) heeft \(\text{rc}_{m} = {\Delta y \over \Delta x} = {3 - 7 \over 1 - 4} = 1\frac{1}{3} \text{.}\)

\(\begin{rcases}l \perp m \text{, dus } \text{rc}_{l} ⋅ \text{rc}_{m} = -1 \\ \text{rc}_{m} = 1\frac{1}{3}\end{rcases} \text{rc}_{l} = -\frac{3}{4}\)

\(\begin{rcases}y = -\frac{3}{4} x + b \\ \text{door } A (4 , 7)\end{rcases} \begin{matrix}7 = -\frac{3}{4} ⋅ 4 + b \\ 7 = -3 + b \\ b = 10\end{matrix}\)
Dus \(l{:}\,y = -\frac{3}{4} x + 10 \text{.}\)

\(\begin{rcases}c{:}\,x^{2} + y^{2} - 8 x - 10 y - 24 = 0 \\ x = 5\end{rcases}\) geeft
\(5^{2} + y^{2} - 8 ⋅ 5 - 10 y - 24 = 0\)
\(1 y^{2} + -10 y + -39 = 0\)
\((y + 3) (y + -13) = 0\)
\(y = -3 ∨ y = 13\)

\(y_{A} > y_{B} \text{,}\) dus \(A (5 , 13)\) en \(B (5 , -3) \text{.}\)

(kwadraatafsplitsen)
\((x - 4)^{2} - 16 + (y - 5)^{2} - 25 - 24 = 0\)
\((x - 4)^{2} + (y - 5)^{2} = 65\)
Dus \(M (4 , 5) \text{.}\)

(voor de lijn door \(B\) en \(M\) geldt)
\(\text{rc}_{\text{BM}} = {\Delta y \over \Delta x} = {5 - -3 \over 4 - 5} = -8 \text{.}\)

\(\begin{rcases}\text{BM} \perp k \text{, dus } \text{rc}_{\text{BM}} ⋅ \text{rc}_{k} = -1 \\ \text{rc}_{\text{BM}} = -8\end{rcases} \text{rc}_{k} = \frac{1}{8}\)

\(\begin{rcases}k{:}\,y = \frac{1}{8} x + b \\ \text{door } B (5 , -3)\end{rcases} \begin{matrix}-3 = \frac{1}{8} ⋅ 5 + b \\ -3 = \frac{5}{8} + b \\ b = -3\frac{5}{8}\end{matrix}\)
Dus \(k{:}\,y = \frac{1}{8} x - 3\frac{5}{8} \text{.}\)

Stel \(y = 3 x + b \text{.}\)
Substitutie in \(x^{2} + y^{2} + 2 x + 4 y - 5 = 0\) geeft
\(x^{2} + (3 x + b)^{2} + 2 x + 4 (3 x + b) - 5 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + 9 x^{2} + 6 b x + b^{2} + 2 x + 12 x + 4 b - 5 = 0\)
\(10 x^{2} + (6 b + 14) x + (b^{2} + 4 b - 5) = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = (6 b + 14)^{2} - 4 ⋅ 10 ⋅ (b^{2} + 4 b - 5)\)
\(D = 36 b^{2} + 168 b + 196 - 40 b^{2} - 160 b + 200\)
\(D = -4 b^{2} + 8 b + 396\)

Oplossen van \(D = 0\) geeft
\(-4 b^{2} + 8 b + 396 = 0\)
\(b^{2} - 2 b - 99 = 0\)
\((b + 9) (b - 11) = 0\)
\(b = -9 ∨ b = 11\)

De vergelijkingen zijn \(y = 3 x - 9\) en \(y = 3 x + 11 \text{.}\)

Stel \(y = a x + 11 \text{.}\)
Substitutie in \(x^{2} + y^{2} - 12 x - 2 y + 29 = 0\) geeft
\(x^{2} + (a x + 11)^{2} - 12 x - 2 (a x + 11) + 29 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + a^{2} x^{2} + 22 a x + 121 - 12 x - 2 a x - 22 + 29 = 0\)
\((1 + a^{2}) x^{2} + (20 a - 12) x + 128 = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = (20 a - 12)^{2} - 4 ⋅ (1 + a^{2}) ⋅ 128\)
\(D = 400 a^{2} - 480 a + 144 - 512 - 512 a^{2}\)
\(D = -112 a^{2} - 480 a - 368\)

Oplossen van \(D = 0\) geeft
\(-112 a^{2} - 480 a - 368 = 0\)
\(-7 a^{2} - 30 a - 23 = 0\)
\(D = (-30)^{2} - 4 ⋅ -7 ⋅ -23 = 256\) dus \(\sqrt{D} = 16\)
\(a = {30 - 16 \over -14} = -1 ∨ a = {30 + 16 \over -14} = -3\frac{2}{7} \text{.}\)

De vergelijkingen zijn \(y = -x + 11\) en \(y = -3\frac{2}{7} x + 11 \text{.}\)