Question

Pair of tangents are drawn from a point $P$ on $x^2+y^2=4$ to $x^2+y^2-20x-20y+199=0$. If $A$ and $B$ are points of contact of these tangents, then the area bounded by the locus of circumcentre of $△PAB$ is (in sq. units)

• A. $\pi$
• B. $\pi$
• C. $4\pi$
• D. $100\pi$

Answer 1

The correct option is A $\pi$

Let $P=(2\cos \theta ,2\sin \theta )$
$x^2+y^2–20x–20y+199=0$
Center $C=(10,10)$


$\because PACB\text{is a cyclic quadrilateral}$
$\therefore$ Circumcircle of $△PAB$ has $PC$ as diameter, so the circumcentre $(h,k)$ is midpoint of $P$ and $C$
$(h,k)=(5+\cos \theta ,5+\sin \theta )$
Therefore, the locus is
$(h-5{)}^2+(k-5{)}^2=1\Rightarrow (x-5{)}^2+(y-5{)}^2=1$
Hence, the required area $=\pi (1{)}^2=\pi \text{sq. units}$