JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Circle

The circle has centre $C(1,3)$ and radius $2$, and $PC=\sqrt{(1+1{)}^2+(3-1{)}^2}=2\sqrt{2}$, so the tangent length is $PA=PB=\sqrt{P{C}^2-4}=2$.
In the right triangle $PAC$, $\tan \angle APC=\frac{CA}{PA}=1$, so $\angle APC=45^{\circ }$ and $\angle APB=90^{\circ }$; with $PA=PB=2$ this gives $AB=2\sqrt{2}$.
A chord of length $2\sqrt{2}$ in a circle of radius $2$ subtends $90^{\circ }$ at the centre, so $\angle ACB=90^{\circ }$. Taking $AD=AB=2\sqrt{2}$ with $D$ distinct from $B$ places $D$ on the other side of $CA$ with $\angle ACD=90^{\circ }$, hence $\angle BCD=180^{\circ }$ and $BD$ is a diameter of length $4$.
Then $A{B}^2+A{D}^2=8+8=16=B{D}^2$, so the triangle is right angled at $A$ and

The area of triangle $ABD$ is $4$ square units.