Raaklijnen aan cirkels

Kwadraatafsplitsen geeft \(x^{2} + (y + 2)^{2} = 17\)
Dus \(M (0 , -2)\) en \(r = \sqrt{17} \text{.}\)

De lijn \(m\) door \(M\) en \(A\) heeft \(\text{rc}_{m} = {\Delta y \over \Delta x} = {-2 - -1 \over 0 - 4} = \frac{1}{4} \text{.}\)

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

\(\begin{rcases}y = -4 x + b \\ \text{door } A (4 , -1)\end{rcases} \begin{matrix}-1 = -4 ⋅ 4 + b \\ -1 = -16 + b \\ b = 15\end{matrix}\)
Dus \(l{:}\,y = -4 x + 15 \text{.}\)

\(\begin{rcases}c{:}\,x^{2} + y^{2} - 6 x - 10 y - 31 = 0 \\ x = 7\end{rcases}\) geeft
\(7^{2} + y^{2} - 6 ⋅ 7 - 10 y - 31 = 0\)
\(1 y^{2} + -10 y + -24 = 0\)
\((y + 2) (y + -12) = 0\)
\(y = -2 ∨ y = 12\)

\(y_{A} > y_{B} \text{,}\) dus \(A (7 , 12)\) en \(B (7 , -2) \text{.}\)

(kwadraatafsplitsen)
\((x - 3)^{2} - 9 + (y - 5)^{2} - 25 - 31 = 0\)
\((x - 3)^{2} + (y - 5)^{2} = 65\)
Dus \(M (3 , 5) \text{.}\)

(voor de lijn door \(B\) en \(M\) geldt)
\(\text{rc}_{\text{BM}} = {\Delta y \over \Delta x} = {5 - -2 \over 3 - 7} = -1\frac{3}{4} \text{.}\)

\(\begin{rcases}\text{BM} \perp l \text{, dus } \text{rc}_{\text{BM}} ⋅ \text{rc}_{l} = -1 \\ \text{rc}_{\text{BM}} = -1\frac{3}{4}\end{rcases} \text{rc}_{l} = \frac{4}{7}\)

\(\begin{rcases}l{:}\,y = \frac{4}{7} x + b \\ \text{door } B (7 , -2)\end{rcases} \begin{matrix}-2 = \frac{4}{7} ⋅ 7 + b \\ -2 = 4 + b \\ b = -6\end{matrix}\)
Dus \(l{:}\,y = \frac{4}{7} x - 6 \text{.}\)

Stel \(y = \frac{1}{2} x + b \text{.}\)
Substitutie in \(x^{2} + y^{2} + 2 x + 8 y + 12 = 0\) geeft
\(x^{2} + (\frac{1}{2} x + b)^{2} + 2 x + 8 (\frac{1}{2} x + b) + 12 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + \frac{1}{4} x^{2} + b x + b^{2} + 2 x + 4 x + 8 b + 12 = 0\)
\(1\frac{1}{4} x^{2} + (b + 6) x + (b^{2} + 8 b + 12) = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = (b + 6)^{2} - 4 ⋅ 1\frac{1}{4} ⋅ (b^{2} + 8 b + 12)\)
\(D = b^{2} + 12 b + 36 - 5 b^{2} - 40 b - 60\)
\(D = -4 b^{2} - 28 b - 24\)

Oplossen van \(D = 0\) geeft
\(-4 b^{2} - 28 b - 24 = 0\)
\(b^{2} + 7 b + 6 = 0\)
\((b + 6) (b + 1) = 0\)
\(b = -6 ∨ b = -1\)

De vergelijkingen zijn \(y = \frac{1}{2} x - 6\) en \(y = \frac{1}{2} x - 1 \text{.}\)

Stel \(y = a x - 3 \text{.}\)
Substitutie in \(x^{2} + y^{2} - 10 x + 8 y + 23 = 0\) geeft
\(x^{2} + (a x - 3)^{2} - 10 x + 8 (a x - 3) + 23 = 0 \text{.}\)

Omschrijven naar de vorm \(A x^{2} + B x + C\) geeft
\(x^{2} + a^{2} x^{2} - 6 a x + 9 - 10 x + 8 a x - 24 + 23 = 0\)
\((1 + a^{2}) x^{2} + (2 a - 10) x + 8 = 0\)

De discriminant is gelijk aan
\(D = B^{2} - 4 A C\)
\(D = (2 a - 10)^{2} - 4 ⋅ (1 + a^{2}) ⋅ 8\)
\(D = 4 a^{2} - 40 a + 100 - 32 - 32 a^{2}\)
\(D = -28 a^{2} - 40 a + 68\)

Oplossen van \(D = 0\) geeft
\(-28 a^{2} - 40 a + 68 = 0\)
\(-7 a^{2} - 10 a + 17 = 0\)
\(D = (-10)^{2} - 4 ⋅ -7 ⋅ 17 = 576\) dus \(\sqrt{D} = 24\)
\(a = {10 - 24 \over -14} = 1 ∨ a = {10 + 24 \over -14} = -2\frac{3}{7} \text{.}\)

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