Raaklijnen aan cirkels

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

De lijn \(m\) door \(M\) en \(A\) heeft \(\text{rc}_{m} = {\Delta y \over \Delta x} = {3 - 5 \over 4 - 8} = \frac{1}{2} \text{.}\)

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

\(\begin{rcases}y = -2 x + b \\ \text{door } A (8 , 5)\end{rcases} \begin{matrix}5 = -2 ⋅ 8 + b \\ 5 = -16 + b \\ b = 21\end{matrix}\)
Dus \(l{:}\,y = -2 x + 21 \text{.}\)

\(\begin{rcases}c{:}\,x^{2} + y^{2} - 4 y - 9 = 0 \\ x = 2\end{rcases}\) geeft
\(2^{2} + y^{2} + 0 ⋅ 2 - 4 y - 9 = 0\)
\(1 y^{2} + -4 y + -5 = 0\)
\((y + 1) (y + -5) = 0\)
\(y = -1 ∨ y = 5\)

\(y_{A} > y_{B} \text{,}\) dus \(A (2 , 5)\) en \(B (2 , -1) \text{.}\)

(kwadraatafsplitsen)
\(x^{2} + (y - 2)^{2} - 4 - 9 = 0\)
\(x^{2} + (y - 2)^{2} = 13\)
Dus \(M (0 , 2) \text{.}\)

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

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

\(\begin{rcases}k{:}\,y = \frac{2}{3} x + b \\ \text{door } B (2 , -1)\end{rcases} \begin{matrix}-1 = \frac{2}{3} ⋅ 2 + b \\ -1 = 1\frac{1}{3} + b \\ b = -2\frac{1}{3}\end{matrix}\)
Dus \(k{:}\,y = \frac{2}{3} x - 2\frac{1}{3} \text{.}\)

Stel \(y = \frac{1}{2} x + b \text{.}\)
Substitutie in \(x^{2} + y^{2} + 6 x - 12 y + 40 = 0\) geeft
\(x^{2} + (\frac{1}{2} x + b)^{2} + 6 x - 12 (\frac{1}{2} x + b) + 40 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + \frac{1}{4} x^{2} + b x + b^{2} + 6 x - 6 x - 12 b + 40 = 0\)
\(1\frac{1}{4} x^{2} + b x + (b^{2} - 12 b + 40) = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = b^{2} - 4 ⋅ 1\frac{1}{4} ⋅ (b^{2} - 12 b + 40)\)
\(D = b^{2} + 0 - 5 b^{2} + 60 b - 200\)
\(D = -4 b^{2} + 60 b - 200\)

Oplossen van \(D = 0\) geeft
\(-4 b^{2} + 60 b - 200 = 0\)
\(b^{2} - 15 b + 50 = 0\)
\((b - 5) (b - 10) = 0\)
\(b = 5 ∨ b = 10\)

De vergelijkingen zijn \(y = \frac{1}{2} x + 5\) en \(y = \frac{1}{2} x + 10 \text{.}\)

Stel \(y = a x + 1 \text{.}\)
Substitutie in \(x^{2} + y^{2} - 8 x - 6 y + 23 = 0\) geeft
\(x^{2} + (a x + 1)^{2} - 8 x - 6 (a x + 1) + 23 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + a^{2} x^{2} + 2 a x + 1 - 8 x - 6 a x - 6 + 23 = 0\)
\((1 + a^{2}) x^{2} + (-4 a - 8) x + 18 = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = (-4 a - 8)^{2} - 4 ⋅ (1 + a^{2}) ⋅ 18\)
\(D = 16 a^{2} + 64 a + 64 - 72 - 72 a^{2}\)
\(D = -56 a^{2} + 64 a - 8\)

Oplossen van \(D = 0\) geeft
\(-56 a^{2} + 64 a - 8 = 0\)
\(-7 a^{2} + 8 a - 1 = 0\)
\(D = 8^{2} - 4 ⋅ -7 ⋅ -1 = 36\) dus \(\sqrt{D} = 6\)
\(a = {-8 - 6 \over -14} = 1 ∨ a = {-8 + 6 \over -14} = \frac{1}{7} \text{.}\)

De vergelijkingen zijn \(y = x + 1\) en \(y = \frac{1}{7} x + 1 \text{.}\)