Co - Axial System of Circles

A system of circles or family of circles, every pair of which have the same radical axis are called co-axial circles

1. The equation of family of co-axial circles when the equation of radical axis and one circle are given

$\mathrm{L}=\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$

$\mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$

Then equation of co-axial circle is $\mathrm{S}+\lambda \mathrm{L}=0$

2. The equation of co-axial system of a circles where the equation of any two circles of the system are

${\mathrm{S}}_1={\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$

${\mathrm{S}}_2={\mathrm{x}}^2+{\mathrm{y}}^2+2{\mathrm{g}}_1\mathrm{x}+2{\mathrm{f}}_1\mathrm{y}+{\mathrm{c}}_1=0$

respectively is ${\mathrm{S}}_1+\lambda ({\mathrm{S}}_1-{\mathrm{S}}_2)=0$

and ${\mathrm{S}}_2+\lambda ({\mathrm{S}}_1-{\mathrm{S}}_2)=0(\lambda \ne -1)$

$S_1+\lambda S_2=0(\lambda \ne -1)$

3. The equation of a system of co-axial circles in the simplest form is ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+\mathrm{c}=0$ where $\mathrm{g}$ is variable and $\mathrm{c}$ is a constant

The common radical axis is the $\mathrm{y}$-axis (centre on $\mathrm{x}$-axis)

The equation of system of co-axial circles in the simplest form is $x^2+y^2+2fy+c=0$ where $f$ is a variable and $\mathrm{c}$ is a constant

The common radical axis is the $\mathrm{x}$-axis (centre on $\mathrm{y}$ axis)

Examples

1. Find the equation of the system of circles co-axial with the circles $x^2+y^2+4x+2y+1=0$ and $x^2+y^2-$ $2x+6y-6=0$. Also find the equation of that particular circles whose centre lies on radical axis.

2. If the circumference of the circle $x^2+y^2+8x+8y-b=0$ is bisected by the circle

$x^2+y^2-2x+4y+a=0$ then $a+b$ equal to

(a) $50$

(b) $56$

(c) $-56$

(d) $-34$

3. The equation of the circle passing through the point of intersection of the circles $x^2+y^2-4x-2y=8$ and $x^2+y^2-2x-4y=8$ and the point $(-1,4)$ is

(a) $x^2+y^2+4x+4y-8=0$

(b) $x^2+y^2-3x+4y+8=0$

(c) $x^2+y^2+x+y-8=0$

(d) $x^2+y^2-3x-3y-8=0$

4. If the common chord of the circles $x^2+(y-b{)}^2=16$ and $x^2+y^2=16$ subtends a right angle at the origin then $\mathrm{b}=$

(a) $4$

(b) $4\sqrt{2}$

(c) $-4\sqrt{2}$

(d) $8$