JEE Advanced 2021 Paper 2 Mathematics: Paragraph 3

The circle has centre $C(3,2)$ and radius $r=\sqrt{9+4-9}=2$. The distance $CP=\sqrt{16+9}=5$, so the tangent length is $PA=\sqrt{25-4}=\sqrt{21}$. Triangle $CAP$ is right-angled at $A$; the chord $AB$ is perpendicular to $CP$, and by area, $\frac12 AM\cdot CP=\frac12 CA\cdot PA$, so half the chord is $AM=\frac{2\sqrt{21}}{5}$. Therefore $AB=\frac{4\sqrt{21}}5=3.666\ldots\approx3.67$. A numerical check with coordinates gives the same value.

With $M$ the midpoint of $AB$, the distance from $P$ to the chord is $PM=\frac{PA^2}{PC}=\frac{21}{5}=4.2$, since in the right triangle $CAP$ the foot of the altitude from $A$ divides the hypotenuse so that $PA^2=PM\cdot PC$. The chord length is $AB=\frac{4\sqrt{21}}5$ from the previous part. Hence the area is $\frac12\cdot AB\cdot PM=\frac12\cdot\frac{4\sqrt{21}}{5}\cdot\frac{21}{5}=\frac{42\sqrt{21}}{25}\approx7.6987$. Rounded, the area is $7.70$ square units, which I confirmed numerically.