Question

Let $P$ be any moving point on the circle $S_1:x^2+y^2-2x-1=0.$ A chord of contact is drawn from the point $P$ to the circle $S:x^2+y^2-2x=0.$ If $C$ is the centre and $A,B$ are the points of contact of circle $S,$ then the locus of the circumcentre of $△CAB$ is

• A. $(x-1{)}^2+y^2=\frac{1}{2}$
• B. $x^2+(y-1{)}^2=\frac{1}{2}$
• C. $(x-1{)}^2+y^2=1$
• D. $x^2+(y-1{)}^2=\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $(x-1{)}^2+y^2=\frac{1}{2}$


The two circles are
$S:(x-1{)}^2+y^2=1$
$S_1:(x-1{)}^2+y^2=2$
Radius of $S$ is $1$ and radius of $S_1$ is $\sqrt{2}$ with both $S$ and $S_1$ having centre as $C(1,0).$
$S_1$ is the director circle of $S$.
$\therefore \angle APB=\frac{\pi }{2}$
Also, $\angle ACB=\frac{\pi }{2}$

Now, circumcentre of the right angled isosceles triangle $△CAB$ would lie on the mid-point of $AB$.
So, let the point be $M\equiv (h,k)$
Now, $CM=CB\sin {45}^{\circ }=\frac{1}{\sqrt{2}}$
$\Rightarrow C{M}^2={\left(\frac{1}{\sqrt{2}}\right)}^2$
$\Rightarrow (h-1{)}^2+(k-0{)}^2={\left(\frac{1}{\sqrt{2}}\right)}^2$
Hence, locus of $M$ is $(x-1{)}^2+y^2=\frac{1}{2}.$