Determine the limiting points of the system of circles $\;$ $x^2 + y^2 -3x - 9y + 45 - \lambda \left(x^2 + y^2 -2x -6y + 30\right) = 0$
Given equation of system of circles
$x^2 + y^2 -3x - 9y + 45 - \lambda \left(x^2 + y^2 -2x -6y + 30\right) = 0$ $\;\;\; \cdots \; (1)$
Equation $(1)$ can be written as
$x^2 \left(1 - \lambda\right) + y^2 \left(1 - \lambda\right) + \left(2 \lambda - 3\right) x + \left(6 \lambda - 9\right) y + \left(45 - 30 \lambda\right) = 0$
i.e. $\;$ $x^2 + y^2 + \left(\dfrac{2 \lambda - 3}{1 - \lambda}\right) x + \left(\dfrac{6 \lambda - 9}{1 - \lambda}\right) y + \dfrac{45 - 30 \lambda}{1 - \lambda} = 0$ $\;\;\; \cdots \; (2)$
Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = \dfrac{2 \lambda - 3}{2 - 2 \lambda}, \;\; f = \dfrac{6 \lambda - 9}{2 - 2 \lambda}, \;\; c = \dfrac{45 - 30 \lambda}{1 - \lambda}$
Center of equation $(2)$ $= \left(-g, -f\right) = \left(\dfrac{3 - 2 \lambda}{2 - 2 \lambda}, \dfrac{9 - 6 \lambda}{2 - 2\lambda}\right)$ $\;\;\; \cdots \; (3a)$
Radius of equation $(2) = r = \sqrt{g^2 + f^2 - c}$
i.e. $\;$ $r = \sqrt{\left(\dfrac{2 \lambda - 3}{2 - 2 \lambda}\right)^2 + \left(\dfrac{6 \lambda - 9}{2 - 2 \lambda}\right)^2 - \left(\dfrac{45 - 30 \lambda}{1 - \lambda}\right)}$ $\;\;\; \cdots \; (3b)$
The limiting points are the centers of circles of the system whose radii are zero.
$\therefore \;$ We have from equation $(3b)$
$\left(\dfrac{2 \lambda - 3}{2 - 2 \lambda}\right)^2 + \left(\dfrac{6 \lambda - 9}{2 - 2 \lambda}\right)^2 - \left(\dfrac{45 - 30 \lambda}{1 - \lambda}\right) = 0$
i.e. $\;$ $\dfrac{4 \lambda^2 - 12 \lambda + 9 + 36 \lambda^2 - 108 \lambda + 81}{4 \left(1 - \lambda\right)^2} = \dfrac{15 \left(3 - 2 \lambda\right)}{1 - \lambda}$
i.e. $\;$ $\dfrac{40 \lambda^2 - 120 \lambda + 90}{4 \left(1 - \lambda\right)} = 15 \left(3 - 2 \lambda\right)$ $\;\;\;$ provided $\;$ $1 - \lambda \neq 0$
i.e. $\;$ $20 \lambda^2 - 60 \lambda + 45 = 30 \left(3 - 2 \lambda\right) \left(1 - \lambda\right)$
i.e. $\;$ $4 \lambda^2 - 12 \lambda + 9 = 6 \left(3 - 5 \lambda + 12 \lambda^2\right)$
i.e. $\;$ $8 \lambda^2 - 18 \lambda + 9 = 0$
$\implies$ $\lambda = \dfrac{3}{4}$ $\;$ or $\;$ $\lambda = \dfrac{3}{2}$
Substituting the values of $\lambda$ in equation $(3a)$ gives the limiting points.
When $\;$ $\lambda = \dfrac{3}{4}$, $\;$ the limiting point is
$\left(\dfrac{3 - 2 \times \dfrac{3}{4}}{2 - 2 \times\dfrac{3}{4}}, \; \dfrac{9 - 6 \times \dfrac{3}{4}}{2 - 2 \times \dfrac{3}{4}}\right) = \left(3, 9\right)$
When $\;$ $\lambda = \dfrac{3}{2}$, $\;$ the limiting point is
$\left(\dfrac{3 - 2 \times \dfrac{3}{2}}{2 - 2 \times \dfrac{3}{2}}, \; \dfrac{9 - 6 \times \dfrac{3}{2}}{2 - 2 \times \dfrac{3}{2}}\right) = \left(0, 0\right)$