Here you will learn equation of family of circles with examples.
Letโ begin โ
Family of Circles
(a)ย The equation of the family of circles passing through the point of intersection of two circles \(S_1\) = 0 & \(S_2\) = 0 is :
\(S_1\) + K\(S_2\) = 0ย ย ย (K \(\ne\) -1)
(b)ย The equation of the family of circles passing through the point of intersection of circle S = 0 & a line L = 0 is given by
S + KL = 0.
(c)ย The equation of the family of circles passing through two given points (\(x_1, y_1\)) & (\(x_2, y_2\)) can be written in the form :
(x โ \(x_1\))(x โ \(x_2\)) + (y โ \(y_1\))(y โ \(y_2\)) + K\begin{vmatrix}
x & y & 1 \\
x_1 & y_1 & 1 \\
x_2 & y_2 & 1
\end{vmatrix} = 0 where K is a parameter.
(d)ย The equation of the family of circles touching a fixed line y โ \(y_1\) = m(x โ \(x_1\)) at the fixed point (\(x_1, y_1\)) is \({(x โ x_1)}^2\) + \({(y โ y_1)}^2\) + K [y โ \(y_1\) โ m(x โ \(x_1\))] = 0,where K is a parameter.
(e)ย Family of circles circumscribing a triangle whose sides are given by \(L_1\) = 0 ; \(L_2\) = 0 & \(L_3\) = 0 is given by : \(L_1\)\(L_2\) + \(\lambda\)\(L_2\)\(L_3\) + \(\mu\)ย \(L_3\)\(L_1\) = 0 provided coefficients of xy = 0 & coefficient of \(x^2\) = coefficient of \(y^2\).
(f)ย Family of circles circumscribing a quadrilateral whose sides are given by \(L_1\) = 0, \(L_2\) = 0, \(L_3\) = 0 & \(L_4\) = 0 is \(L_1\)\(L_3\) + \(\lambda\)\(L_2\)\(L_4\) provided coefficients of xy = 0 & coefficient of \(x^2\) = coefficient of \(y^2\).
Example : Find the equation of the circle through the points of intersection of \(x^2 + y^2 โ 1\) = 0, \(x^2 + y^2 โ 2x โ 4y + 1\) = 0 and touching the line x + 2y = 0.
Solution : Family of circles is \(x^2 + y^2 โ 2x โ 4y + 1\) + \(\lambda\)(\(x^2 + y^2 โ 1\)) = 0
(1 + \(\lambda\))\(x^2\) + (1 + \(\lambda\))\(y^2\) โ 2x โ 4y + (1 โ \(\lambda\))) = 0
\(x^2 + y^2 โ {2\over {1 + \lambda}} x โ {4\over {1 + \lambda}}y + {{1 โ \lambda}\over {1 + \lambda}}\) = 0
Centre is (\({1\over {1 + \lambda}}\), \({2\over {1 + \lambda}}\))ย and radius = \(\sqrt{4 + {\lambda}^2}\over |1 + \lambda|\)
Since it touches the line x + 2y = 0, hence
Radius = Perpendicular distance from center to the line
i.e., |\({1\over {1 + \lambda}} + 2{2\over {1 + \lambda}}\over \sqrt{1^2 + 2^2}\)| = \(\sqrt{4 + {\lambda}^2}\over |1 + \lambda|\) \(\implies\) \(\sqrt{5}\) = \(\sqrt{4 + {\lambda}^2}\) \(\implies\) \(\lambda\) = \(\pm\) 1.
\(\lambda\) = -1 cannot be possible in case of circle. So \(\lambda\) = 1.
Hence the equation of the circle is \(x^2 + y^2 โ x โ 2y\) = 0
