Raaklijnen aan cirkels

Kwadraatafsplitsen geeft \((x - 3)^{2} + (y + 5)^{2} = 10\)
Dus \(M (3 , -5)\) en \(r = \sqrt{10} \text{.}\)

De lijn \(m\) door \(M\) en \(A\) heeft \(\text{rc}_{m} = {\Delta y \over \Delta x} = {-5 - -4 \over 3 - 6} = \frac{1}{3} \text{.}\)

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

\(\begin{rcases}y = -3 x + b \\ \text{door } A (6 , -4)\end{rcases} \begin{matrix}-4 = -3 ⋅ 6 + b \\ -4 = -18 + b \\ b = 14\end{matrix}\)
Dus \(l{:}\,y = -3 x + 14 \text{.}\)

\(\begin{rcases}c{:}\,x^{2} + y^{2} + 4 x - 16 = 0 \\ x = 2\end{rcases}\) geeft
\(2^{2} + y^{2} + 4 ⋅ 2 - 16 = 0\)
\(1 y^{2} + 0 y + -4 = 0\)
\(1 y^{2} = 4\)
\(y = -2 ∨ y = 2\)

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

(kwadraatafsplitsen)
\((x + 2)^{2} - 4 + y^{2} - 16 = 0\)
\((x + 2)^{2} + y^{2} = 20\)
Dus \(M (-2 , 0) \text{.}\)

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

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

\(\begin{rcases}l{:}\,y = 2 x + b \\ \text{door } B (2 , -2)\end{rcases} \begin{matrix}-2 = 2 ⋅ 2 + b \\ -2 = 4 + b \\ b = -6\end{matrix}\)
Dus \(l{:}\,y = 2 x - 6 \text{.}\)

Stel \(y = -2 x + b \text{.}\)
Substitutie in \(x^{2} + y^{2} + 2 x - 16 y + 60 = 0\) geeft
\(x^{2} + (-2 x + b)^{2} + 2 x - 16 (-2 x + b) + 60 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + 4 x^{2} - 4 b x + b^{2} + 2 x + 32 x - 16 b + 60 = 0\)
\(5 x^{2} + (-4 b + 34) x + (b^{2} - 16 b + 60) = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = (-4 b + 34)^{2} - 4 ⋅ 5 ⋅ (b^{2} - 16 b + 60)\)
\(D = 16 b^{2} - 272 b + 1\,156 - 20 b^{2} + 320 b - 1\,200\)
\(D = -4 b^{2} + 48 b - 44\)

Oplossen van \(D = 0\) geeft
\(-4 b^{2} + 48 b - 44 = 0\)
\(b^{2} - 12 b + 11 = 0\)
\((b - 1) (b - 11) = 0\)
\(b = 1 ∨ b = 11\)

De vergelijkingen zijn \(y = -2 x + 1\) en \(y = -2 x + 11 \text{.}\)

Stel \(y = a x - 2 \text{.}\)
Substitutie in \(x^{2} + y^{2} - 10 x + 6 y + 21 = 0\) geeft
\(x^{2} + (a x - 2)^{2} - 10 x + 6 (a x - 2) + 21 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + a^{2} x^{2} - 4 a x + 4 - 10 x + 6 a x - 12 + 21 = 0\)
\((1 + a^{2}) x^{2} + (2 a - 10) x + 13 = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = (2 a - 10)^{2} - 4 ⋅ (1 + a^{2}) ⋅ 13\)
\(D = 4 a^{2} - 40 a + 100 - 52 - 52 a^{2}\)
\(D = -48 a^{2} - 40 a + 48\)

Oplossen van \(D = 0\) geeft
\(-48 a^{2} - 40 a + 48 = 0\)
\(-6 a^{2} - 5 a + 6 = 0\)
\(D = (-5)^{2} - 4 ⋅ -6 ⋅ 6 = 169\) dus \(\sqrt{D} = 13\)
\(a = {5 - 13 \over -12} = \frac{2}{3} ∨ a = {5 + 13 \over -12} = -1\frac{1}{2} \text{.}\)

De vergelijkingen zijn \(y = \frac{2}{3} x - 2\) en \(y = -1\frac{1}{2} x - 2 \text{.}\)