Berkas Lingkaran

Persamaan berkas lingkaran \(L_1 + \lambda \:.\: L_2 = 0\)

\begin{equation*}
\begin{split}
x^2 + y^2 + 6x - 2y + 6 + \lambda \:.\: (x^2 + y^2 - 10x + 10y + 34) & = 0  \\\\
x^2 + y^2 + 6x - 2y + 6 + \lambda \:.\: x^2 + \lambda \:.\: y^2 - 10 \:.\: \lambda \:.\: x + 10 \:.\: \lambda \:.\: y + 34 \:.\: \lambda & = 0  \\\\
(1 + \lambda) \:.\: x^2 + (1 + \lambda) \:.\: y^2 + (6 - 10 \lambda)x + (10 \lambda - 2)y + 6 + 34 \lambda & = 0
\end{split}
\end{equation*}

Substitusi titik (1,1) ke berkas lingkaran

\begin{equation*}
\begin{split}
(1 + \lambda) \:.\: x^2 + (1 + \lambda) \:.\: y^2 + (6 - 10 \lambda)x + (10 \lambda - 2)y + 6 + 34 \lambda & = 0 \\\\
(1 + \lambda) \:.\: 1^2 + (1 + \lambda) \:.\: 1^2 + (6 - 10 \lambda) \:.\: 1 + (10 \lambda - 2) \:.\: 1 + 6 + 34 \lambda & = 0 \\\\
1 + \lambda + 1 + \lambda + 6 - 10 \lambda + 10 \lambda - 2 + 6 + 34 \lambda & = 0 \\\\
36 \lambda + 12 & = 0 \\\\
36 \lambda & = -12 \\\\
\lambda & = - \frac 13
\end{split}
\end{equation*}

Persamaan lingkaran

\begin{equation*}
\begin{split}
(1 + \lambda) \:.\: x^2 + (1 + \lambda) \:.\: y^2 + (6 - 10 \lambda)x + (10 \lambda - 2)y + 6 + 34 \lambda & = 0 \\\\
\left(1 - \frac 13 \right) \:.\: x^2 + \left(1 - \frac 13 \right) \:.\: y^2 + \left(6 - 10 \:.\: - \frac 13 \right)x + \left (10 \:.\: - \frac 13 - 2 \right)y + 6 + 34 \:.\: - \frac 13 & = 0 \\\\
\frac 23 x^2 + \frac 23 y^2 + \frac {28}{3} x - \frac {16}{3} y - \frac {16}{3} & = 0 \quad {\color {blue} \times \frac 32}\\\\
x^2 + y^2 + 14 x - 8 y - 8 & = 0
\end{split}
\end{equation*}