The limiting points of the co-axial system containing the two circles $\begin{aligned} & x^2+y^2+2 x-2 y+2=0…
The limiting points of the co-axial system containing the two circles
$\begin{aligned} & x^2+y^2+2 x-2 y+2=0 \text { and } \\ & 25\left(x^2+y^2\right)-10 x-80 y+65=0 \text { are }\end{aligned}$
$(1,-1),(-3,-40)$
$(1,-1),\left(-\frac{1}{5}, \frac{8}{5}\right)$
$(-1,1),\left(\frac{1}{5}, \frac{8}{5}\right)$
$\left(-\frac{1}{5},-\frac{8}{5}\right)$
Solution
Equation of circles are
$x^2+y^2+2 x-2 y+2=0$
and $x^2+y^2-\frac{2}{5} x-\frac{16}{5} y+\frac{13}{5}=0$
$\begin{aligned} & \Rightarrow \quad\left(x^2+y^2+2 x-2 y+2\right) \\ & +\lambda\left[x^2+y^2-\frac{2}{5} x-\frac{16}{5} y+\frac{13}{5}\right]=0 \\ & \Rightarrow \quad(1+\lambda) x^2+(1+\lambda) y^2+2\left(1-\frac{\lambda}{5}\right) x \\ & \quad-2\left(1+\frac{8 \lambda}{5}\right) y+\left(2+\frac{13 \lambda}{5}\right)=0 \\ & \Rightarrow \quad\left(1-\frac{\lambda}{5}\right) \\ & \quad x^2+y^2+2 \frac{1+\lambda}{1+\lambda} x \\ & \quad-\frac{2\left(1+\frac{8 \lambda}{5}\right)}{1+\lambda} y+\frac{2+\frac{13 \lambda}{5}}{1+\lambda}=0\end{aligned}$
At
$g=0 \Rightarrow \frac{1-\frac{\lambda}{5}}{1+\lambda}=0 \Rightarrow \lambda=5$
Then $\quad x^2+y^3-3 y+\frac{5}{2}=0$
$\therefore$ Centre is $\left(0, \frac{3}{2}\right)$.
The limiting points are $(-1,1),\left(\frac{1}{5}, \frac{8}{5}\right)$.