Find the required system of circles

Question

Show that the equation $x^2+y^2-2x-2ay-8=0$ represents for different values of '$a$' a system of circles passing through two fixed points $A$,$B$ on the $x$-axis and find the equation of that circle of the system, the tangents to which at $A$ and $B$ meet on the line $x+2y+5=0$.

Answer 1

$x^2+y^2 -2x - 2ay - 8 = 0\implies (x-1)^2 +(y-a)^2= (\sqrt{a^2+9})^2$. Put $y = 0$, and for any $a$, $(x-1)^2 = 9\implies x-1 = \pm 3\implies x = -2,4$. The $2$ fixed points are $(-2,0), (4,0)$.

Answer 2

$\begin{align} x^2+y^2-2x-2ay-8= 0 &\iff (x-1)^2-1 +(y-a)^2-a^2-8 = 0 \\ &\iff (x-1)^2+(y-a)^2 = 9+a^2 \end{align}$

Which is the equation of a circle centered in $\Omega(1, a)$ with radius $\sqrt{9+a^2}$

Each of those circles intersects the $x$-axis two times, because $(x-1)^2 = 9$ has two solutions : $x=4$ and $x=-2$.

Thus we have $A(-2,0)$ and $B(4, 0)$. Now find the equation of the tangent at $A$ by finding an orthogonal vector to $\overrightarrow{\Omega A}$, do the same with $B$, and find the intersection of the tangents.