Raaklijnen aan cirkels

Kwadraatafsplitsen geeft \((x + 1)^{2} + (y - 3)^{2} = 20\)
Dus \(M (-1 , 3)\) en \(r = \sqrt{20} \text{.}\)

De lijn \(m\) door \(M\) en \(A\) heeft \(\text{rc}_{m} = {\Delta y \over \Delta x} = {3 - 7 \over -1 - 1} = 2 \text{.}\)

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

\(\begin{rcases}y = -\frac{1}{2} x + b \\ \text{door } A (1 , 7)\end{rcases} \begin{matrix}7 = -\frac{1}{2} ⋅ 1 + b \\ 7 = -\frac{1}{2} + b \\ b = 7\frac{1}{2}\end{matrix}\)
Dus \(l{:}\,y = -\frac{1}{2} x + 7\frac{1}{2} \text{.}\)

\(\begin{rcases}c{:}\,x^{2} + y^{2} + 6 x + 4 y - 32 = 0 \\ x = 0\end{rcases}\) geeft
\(0^{2} + y^{2} + 6 ⋅ 0 + 4 y - 32 = 0\)
\(1 y^{2} + 4 y + -32 = 0\)
\((y + 8) (y + -4) = 0\)
\(y = -8 ∨ y = 4\)

\(y_{A} < y_{B} \text{,}\) dus \(A (0 , -8)\) en \(B (0 , 4) \text{.}\)

(kwadraatafsplitsen)
\((x + 3)^{2} - 9 + (y + 2)^{2} - 4 - 32 = 0\)
\((x + 3)^{2} + (y + 2)^{2} = 45\)
Dus \(M (-3 , -2) \text{.}\)

(voor de lijn door \(A\) en \(M\) geldt)
\(\text{rc}_{\text{AM}} = {\Delta y \over \Delta x} = {-2 - -8 \over -3 - 0} = -2 \text{.}\)

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

\(\begin{rcases}k{:}\,y = \frac{1}{2} x + b \\ \text{door } A (0 , -8)\end{rcases} \begin{matrix}-8 = \frac{1}{2} ⋅ 0 + b \\ -8 = 0 + b \\ b = -8\end{matrix}\)
Dus \(k{:}\,y = \frac{1}{2} x - 8 \text{.}\)

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

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

Stel \(y = a x - 5 \text{.}\)
Substitutie in \(x^{2} + y^{2} - 6 x + 8 y + 23 = 0\) geeft
\(x^{2} + (a x - 5)^{2} - 6 x + 8 (a x - 5) + 23 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + a^{2} x^{2} - 10 a x + 25 - 6 x + 8 a x - 40 + 23 = 0\)
\((1 + a^{2}) x^{2} + (-2 a - 6) x + 8 = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = (-2 a - 6)^{2} - 4 ⋅ (1 + a^{2}) ⋅ 8\)
\(D = 4 a^{2} + 24 a + 36 - 32 - 32 a^{2}\)
\(D = -28 a^{2} + 24 a + 4\)

Oplossen van \(D = 0\) geeft
\(-28 a^{2} + 24 a + 4 = 0\)
\(-7 a^{2} + 6 a + 1 = 0\)
\(D = 6^{2} - 4 ⋅ -7 ⋅ 1 = 64\) dus \(\sqrt{D} = 8\)
\(a = {-6 - 8 \over -14} = 1 ∨ a = {-6 + 8 \over -14} = -\frac{1}{7} \text{.}\)

De vergelijkingen zijn \(y = x - 5\) en \(y = -\frac{1}{7} x - 5 \text{.}\)