Raaklijnen aan cirkels

Kwadraatafsplitsen geeft \(x^{2} + (y + 2)^{2} = 5\)
Dus \(M (0 , -2)\) en \(r = \sqrt{5} \text{.}\)

De lijn \(m\) door \(M\) en \(A\) heeft \(\text{rc}_{m} = {\Delta y \over \Delta x} = {-2 - 0 \over 0 - 1} = 2 \text{.}\)

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

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

\(\begin{rcases}c{:}\,x^{2} + y^{2} + 8 y + 14 = 0 \\ x = 1\end{rcases}\) geeft
\(1^{2} + y^{2} + 0 ⋅ 1 + 8 y + 14 = 0\)
\(1 y^{2} + 8 y + 15 = 0\)
\((y + 5) (y + 3) = 0\)
\(y = -5 ∨ y = -3\)

\(y_{A} < y_{B} \text{,}\) dus \(A (1 , -5)\) en \(B (1 , -3) \text{.}\)

(kwadraatafsplitsen)
\(x^{2} + (y + 4)^{2} - 16 + 14 = 0\)
\(x^{2} + (y + 4)^{2} = 2\)
Dus \(M (0 , -4) \text{.}\)

(voor de lijn door \(B\) en \(M\) geldt)
\(\text{rc}_{\text{BM}} = {\Delta y \over \Delta x} = {-4 - -3 \over 0 - 1} = 1 \text{.}\)

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

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

Stel \(y = 2 x + b \text{.}\)
Substitutie in \(x^{2} + y^{2} + 2 x + 22 y + 117 = 0\) geeft
\(x^{2} + (2 x + b)^{2} + 2 x + 22 (2 x + b) + 117 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + 4 x^{2} + 4 b x + b^{2} + 2 x + 44 x + 22 b + 117 = 0\)
\(5 x^{2} + (4 b + 46) x + (b^{2} + 22 b + 117) = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = (4 b + 46)^{2} - 4 ⋅ 5 ⋅ (b^{2} + 22 b + 117)\)
\(D = 16 b^{2} + 368 b + 2\,116 - 20 b^{2} - 440 b - 2\,340\)
\(D = -4 b^{2} - 72 b - 224\)

Oplossen van \(D = 0\) geeft
\(-4 b^{2} - 72 b - 224 = 0\)
\(b^{2} + 18 b + 56 = 0\)
\((b + 14) (b + 4) = 0\)
\(b = -14 ∨ b = -4\)

De vergelijkingen zijn \(y = 2 x - 14\) en \(y = 2 x - 4 \text{.}\)

Stel \(y = a x - 5 \text{.}\)
Substitutie in \(x^{2} + y^{2} - 8 x + 6 y + 15 = 0\) geeft
\(x^{2} + (a x - 5)^{2} - 8 x + 6 (a x - 5) + 15 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + a^{2} x^{2} - 10 a x + 25 - 8 x + 6 a x - 30 + 15 = 0\)
\((1 + a^{2}) x^{2} + (-4 a - 8) x + 10 = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = (-4 a - 8)^{2} - 4 ⋅ (1 + a^{2}) ⋅ 10\)
\(D = 16 a^{2} + 64 a + 64 - 40 - 40 a^{2}\)
\(D = -24 a^{2} + 64 a + 24\)

Oplossen van \(D = 0\) geeft
\(-24 a^{2} + 64 a + 24 = 0\)
\(-3 a^{2} + 8 a + 3 = 0\)
\(D = 8^{2} - 4 ⋅ -3 ⋅ 3 = 100\) dus \(\sqrt{D} = 10\)
\(a = {-8 - 10 \over -6} = 3 ∨ a = {-8 + 10 \over -6} = -\frac{1}{3} \text{.}\)

De vergelijkingen zijn \(y = 3 x - 5\) en \(y = -\frac{1}{3} x - 5 \text{.}\)