Question

$x^2+y^2-2x-2ay-8=0,$ a is a variable. Equation of a circle $C$ of this family, tangents to which at these fixed points intersects on the
line $x+2y+5=0$ is

• A. $x^2+y^2-2x-8y-8=0$
• B. $x^2+y^2-2x+6y-8=0$
• C. $x^2+y^2-2x+8y-8=0$
• D. $x^2+y^2-2x-6y-8=0$

Answer 1

The correct option is D $x^2+y^2-2x-6y-8=0$

Equation of the given circle can be written as
$(x^2+y^2-2x-8)-2a\left(y\right)=0$
which represents a family of circles passing through the intersection of the circle $x^2+y^2-2x-8=0$ and the line $y=0.$
The circle and the line intersect at the points $P(-2,0)$ and $Q(4,0)$.
Let the tangents at $P$ and $Q$ to a member of this family intersect at $(h,k),$

then $PQ$ is the chord of contact of $(h,k)$ and its equation is
$hx+ky-(x+h)-a(y+k)-8=0$ $\Rightarrow x(h-1)+y(k-a)-(h+ak+8)=0$
Comparing this with equation $y=0$ of $PQ,$ we get

$h=1,$ $h+ak+8=0,$
Since $(h,k)$ lies on the given line

$x+2y+5=0$ $1+2k+5=0\Rightarrow k=-3$ and $1-3a+8=0\Rightarrow a=3$
Hence the equation of the required member $C$ of the family is

$x^2+y^2-2x-6y-8=0$