# 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