Question

One of the limiting points of the coaxial system of circles containing the circle $x^2+y^2-6x-6y+4=0$ and $x^2+y^2-2x-4y+3=0$ is:

Answer 1

$x^2+y^2-6x-6y+4=0$

$x^2+y^2-2x-4y+3=0$

The equation of family of coaxial system of circles having $2$ members is given by

$(x^2+y^2-6x-6y+4)+\lambda (-4x-2y+1)=0$

$\Rightarrow x^2+y^2-2x(3+2\lambda )-2y(3+\lambda )+(4+\lambda )=0$ ……$\left(1\right)$

Centre of $\left(1\right)$ $(3+2\lambda ,3+\lambda )$ & radius of $\left(1\right)$

$\sqrt{(3+2\lambda {)}^2+(3+\lambda {)}^2-(4+\lambda )}=R$

For limiting points radius

$R=0$

$\Rightarrow 5{\lambda }^2+17\lambda +14=0$

$\Rightarrow \lambda =-2,\frac{-7}{5}$

$\therefore$ Limiting points are $(-1,1)$ & $(\frac{1}{5},\frac{8}{5})$.