Question

$x^2+y^2-2x-2ay-8=0,$ a is a variable. If the chord joining the fixed points subtends an angle $\theta$ at the centre of the circle $C,$ then $\theta =$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (D)

$\frac{\pi }{2}$

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
$hx<+ky-(x+h)-a(y+k)-8=0$ or or $x(h-1)+y(k-a)-(h+ak+8)=0$
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 $x^2+y^2-2x-6y-8=0$
Centre of the circle $C$ is $R(1,3)$
$\therefore$ Slope of $PR=1$ and of $QR=-1\Rightarrow \theta =\frac{\pi }{2}$