Afstand tussen punten, lijnen en cirkels

\(M (-4 , -2)\) en \(r = \sqrt{2} \text{.}\)

De lijn \(n\) gaat door \(M\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,4 x - 2 y = c \\ M (-4 , -2)\end{rcases} c = 4 ⋅ -4 - 2 ⋅ -2 = -12\)
Dus \(n{:}\,4 x - 2 y = -12 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}2 x + 4 y = -1 \\ 4 x - 2 y = -12\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}4 x + 8 y = -2 \\ 4 x - 2 y = -12\end{cases}\)
Aftrekken geeft \(10 y = 10\) dus \(y = 1 \text{.}\)

\(\begin{rcases}2 x + 4 y = -1 \\ y = 1\end{rcases} \begin{matrix}2 x + 4 ⋅ 1 = -1 \\ x = -2\frac{1}{2}\end{matrix}\)
Dus \(S (-2\frac{1}{2} , 1) \text{.}\)

\(d(M , l) = d(M , S) = \sqrt{(-4 - -2\frac{1}{2})^{2} + (-2 - 1)^{2}} = \sqrt{11\frac{1}{4}} \text{.}\)

\(d(c , l) = d(M , l) - r = \sqrt{11\frac{1}{4}} - \sqrt{2} \text{.}\)

Kwadraatafsplitsen geeft \((x + 5)^{2} + (y + 4)^{2} = 9\)
Dus \(M (-5 , -4)\) en \(r = \sqrt{9} = 3 \text{.}\)

\(d(M , A) = \sqrt{(-5 - -1)^{2} + (-4 - 0)^{2}} = \sqrt{32} \text{.}\)

Er geldt \(d(M , A) > r \text{,}\) dus \(d(c , A) = d(M , A) - r = \sqrt{32} - 3 \text{.}\)

De lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{.}\)
\(\begin{rcases}n{:}\,-2 x - y = c \\ A (4 , 3)\end{rcases} c = -2 ⋅ 4 - 1 ⋅ 3 = -11\)
Dus \(n{:}\,-2 x - y = -11 \text{.}\)

\(l\) en \(n\) snijden geeft het punt \(S \text{.}\)
\(\begin{cases}x - 2 y = 3 \\ -2 x - y = -11\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}2 x - 4 y = 6 \\ -2 x - y = -11\end{cases}\)
Optellen geeft \(-5 y = -5\) dus \(y = 1 \text{.}\)

\(\begin{rcases}x - 2 y = 3 \\ y = 1\end{rcases} \begin{matrix}x - 2 ⋅ 1 = 3 \\ x = 5\end{matrix}\)
Dus \(S (5 , 1) \text{.}\)

\(d(A , l) = d(A , S) = \sqrt{(4 - 5)^{2} + (3 - 1)^{2}} = \sqrt{5} \text{.}\)

\(M (3 , 5)\) en \(r = \sqrt{27} \text{.}\)

\(d(A , M) = \sqrt{(-4 - 3)^{2} + (-3 - 5)^{2}} = \sqrt{113}\)

\(d(A , c) = d(A , M) - r = \sqrt{113} - \sqrt{27} = 5{,}433...\)

(lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{)}\)
\(\begin{rcases}n{:}\,4 x - 2 y = c \\ A (-4 , -3)\end{rcases} c = 4 ⋅ -4 - 2 ⋅ -3 = -10\)

(\(l\) en \(n\) snijden geeft het punt \(S \text{)}\)
\(\begin{cases}2 x + 4 y = -45 \\ 4 x - 2 y = -10\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}4 x + 8 y = -90 \\ 4 x - 2 y = -10\end{cases}\)
Aftrekken geeft \(10 y = -80\) dus \(y = -8 \text{.}\)

\(\begin{rcases}2 x + 4 y = -45 \\ y = -8\end{rcases} \begin{matrix}2 x + 4 ⋅ -8 = -45 \\ x = -6\frac{1}{2}\end{matrix}\)
Dus \(S (-6\frac{1}{2} , -8) \text{.}\)

\(d(A , l) = d(A , S) = \sqrt{(-4 - -6\frac{1}{2})^{2} + (-3 - -8)^{2}} = \sqrt{31\frac{1}{4}} = 5{,}590...\)

\(d(A , c) < d(A , l) \text{,}\) dus \(A\) ligt dichter bij \(c\) dan bij \(l \text{.}\)

\(x^{2} + y^{2} - 6 x - 2 y + 3 = 0\)
\((x - 3)^{2} - 9 + (y - 1)^{2} - 1 + 3 = 0\)
\((x - 3)^{2} + (y - 1)^{2} = 7\)

\(M (3 , 1)\) en \(r = \sqrt{7} \text{.}\)

\(d(A , M) = \sqrt{(0 - 3)^{2} + (-4 - 1)^{2}} = \sqrt{34}\)

\(d(A , c) = d(A , M) - r = \sqrt{34} - \sqrt{7} = 3{,}185...\)

(lijn \(n\) gaat door \(A\) en staat loodrecht op \(l \text{)}\)
\(\begin{rcases}n{:}\,2 x - 4 y = c \\ A (0 , -4)\end{rcases} c = 2 ⋅ 0 - 4 ⋅ -4 = 16\)

(\(l\) en \(n\) snijden geeft het punt \(S \text{)}\)
\(\begin{cases}4 x + 2 y = -33 \\ 2 x - 4 y = 16\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}4 x + 2 y = -33 \\ 4 x - 8 y = 32\end{cases}\)
Aftrekken geeft \(10 y = -65\) dus \(y = -6\frac{1}{2} \text{.}\)

\(\begin{rcases}4 x + 2 y = -33 \\ y = -6\frac{1}{2}\end{rcases} \begin{matrix}4 x + 2 ⋅ -6\frac{1}{2} = -33 \\ x = -5\end{matrix}\)
Dus \(S (-5 , -6\frac{1}{2}) \text{.}\)

\(d(A , l) = d(A , S) = \sqrt{(0 - -5)^{2} + (-4 - -6\frac{1}{2})^{2}} = \sqrt{31\frac{1}{4}} = 5{,}590...\)

\(d(A , c) < d(A , l) \text{,}\) dus \(A\) ligt dichter bij \(c\) dan bij \(l \text{.}\)

Kwadraatafsplitsen geeft \((x + 3)^{2} + (y - 7)^{2} = 9\)
Dus \(M_{1} (-3 , 7)\) en \(r_{1} = \sqrt{9} \text{.}\)

Het middelpunt van cirkel \(c_{2}\) is \(M_{2} (3 , -2) \text{,}\) dus
\(d(M_{1} , M_{2}) = \sqrt{(-3 - 3)^{2} + (7 - -2)^{2}} = \sqrt{117} \text{.}\)

Er geldt \(r_{2} = \sqrt{12} \text{,}\) dus
\(d(c_{1} , c_{2}) = d(M_{1} , M_{2}) - r_{1} - r_{2} = \sqrt{117} - \sqrt{9} - \sqrt{12} \text{.}\)

\(d(A , B) = \sqrt{(-3 - -4)^{2} + (-1 - -3)^{2}} = \sqrt{1 + 4} = \sqrt{5} \text{.}\)

\(M (2 , 4)\) en \(r = \sqrt{39} \text{.}\)

\(d(M , A) = \sqrt{(7 - 2)^{2} + (8 - 4)^{2}} = \sqrt{41} \text{.}\)

\(d(M , A) > r \text{,}\) dus \(A\) ligt buiten \(c \text{.}\)