Family of Circles – Equations & Examples

Here you will learn equation of family of circles with examples.

Let’ begin –

Family of Circles

(a)  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\(S_2\) = 0      (K \(\ne\) -1)

(b)  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)  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 – \(x_1\))(x – \(x_2\)) + (y – \(y_1\))(y – \(y_2\)) + 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)  The equation of the family of circles touching a fixed line y – \(y_1\) = m(x – \(x_1\)) at the fixed point (\(x_1, y_1\)) is \({(x – x_1)}^2\) + \({(y – y_1)}^2\) + K [y – \(y_1\) – m(x – \(x_1\))] = 0,where K is a parameter.

(e)  Family of circles circumscribing a triangle whose sides are given by \(L_1\) = 0 ; \(L_2\) = 0 & \(L_3\) = 0 is given by : \(L_1\)\(L_2\) + \(\lambda\)\(L_2\)\(L_3\) + \(\mu\)  \(L_3\)\(L_1\) = 0 provided coefficients of xy = 0 & coefficient of \(x^2\) = coefficient of \(y^2\).

(f)  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 \(x^2 + y^2 – 1\) = 0, \(x^2 + y^2 – 2x – 4y + 1\) = 0 and touching the line x + 2y = 0.

Solution : Family of circles is \(x^2 + y^2 – 2x – 4y + 1\) + \(\lambda\)(\(x^2 + y^2 – 1\)) = 0

(1 + \(\lambda\))\(x^2\) + (1 + \(\lambda\))\(y^2\) – 2x – 4y + (1 – \(\lambda\))) = 0

\(x^2 + y^2 – {2\over {1 + \lambda}} x – {4\over {1 + \lambda}}y + {{1 – \lambda}\over {1 + \lambda}}\) = 0

Centre is (\({1\over {1 + \lambda}}\), \({2\over {1 + \lambda}}\))  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 \(x^2 + y^2 – x – 2y\) = 0

Leave a Comment

Your email address will not be published.