Question

Prove that the equation $x^2+y^2-2x-2ay-8=0$ represents a family of circles passing through two fixed points, say $P$ and $Q$. Choose the member of the family tangents to which at $P$ and $Q$ intersect on the line $x+2y+5=0$.

Answer 1

The given circle is
$(x^2+y^2-2x-8)-2ay=0.....\left(1\right)$
It is of the form $S+\lambda P=0$ and hence represents a family of circles passing through the points of intersection $S=0$ and $P=0$.
i.e. $x^2+y^2-2x-8=0$ and $y=0$.
$\therefore P$ and $Q$ are $(-2,0)$ and $(4,0)$.
Equation of $PQ$ is $y=0.....\left(2\right)$
If tangents at $P$ and $Q$ intersect at $R(h,k)$ then $PQ$ is chord of contact of $(h,k)$ whose equation is $x.h+y.k-(x+h)-a(y+k)-8=0$
or $x(h-1)+y(k-a)-(h+ak+8)=0.....\left(3\right)$
Comparing $\left(3\right)$ and $\left(2\right),h-1=0,h+ak+8=0,k-a=$ any constant,
$\therefore h=1$. If $(h,k)$ lies on $x+2y+5=0$
$\therefore h+2k+5=0\therefore k=-3$. Putting for $(h,k)$ in $h+ak+8=0$, we get $a=3$.
Hence the required member of the family of circles is $x^2+y^2-2x-6y-8=0$. by $\left(1\right)$.