Question

The locus of the foot of the normal drawn from any point $P(\alpha ,\beta )$ to the family of circles $x^2+y^2-2gx+c=0$, where $g$ is a parameter, is

• A. $(x^2+y^2+c)(y-\beta )=2(y\alpha -x\beta )x$
• B. $(x^2+y^2+c)(x-\beta )=2(y\alpha -x\beta )x$
• C. $(x^2+y^2+c)(y-\beta )=2(x\alpha -y\beta )x$
• D. none of these

Answer 1

The correct option is A $(x^2+y^2+c)(y-\beta )=2(y\alpha -x\beta )x$

Let $Q(h,k)$ be the foot of the normal drawn from $P(\alpha ,\beta )$ to the given circle.

Then, $P,Q$ and $C$ are collinear points.
$\therefore \left|\begin{array}{ccccc}\alpha & & \beta & & 1\\ h& & k& & 1\\ g& & 0& & 1\end{array}\right|=0$
$\Rightarrow \alpha k+\beta g-\beta h-gk=0$

$\Rightarrow g(\beta -k)=\beta h-\alpha k$ .......(i)
Also, $CQ$ $=$ Radius
$\Rightarrow (g-h{)}^2+k^2=g^2-c$
$\Rightarrow -2gh+h^2+k^2=-c$

$\Rightarrow g=\frac{h^2+k^2+c}{2h}$ .......(ii)
From (i) and (ii), we get

$\frac{h^2+k^2+c}{2h}=\frac{\beta h-\alpha k}{\beta -k}$
Hence, the locus of $(h,k)$ is
$(x^2+y^2+c)(\beta -y)=2(\beta x-\alpha y)x$
Thus, option $\left(a\right)$ is correct.