$A B$ is the chord of the circle $x^{2} + y^{2} = r^{2}$ whose mid-point is $C (a, b) . O$ is the origin and $P$ is the point of intersection of the tangents to the circle at the extremities of the chord $A B$ then

Area of

Δ P A B = \frac{(r^{2} - a^{2} - b^{2})^{\frac{3}{2}}}{\sqrt{a^{2} + b^{2}}}

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Area of

Δ P A B = \frac{(r^{2} + a^{2} + b^{2})^{\frac{3}{2}}}{\sqrt{a^{2} + b^{2}}}

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Area of

Δ O A B = \sqrt{(a^{2} + b^{2}) (r^{2} - a^{2} - b^{2})}

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Area of

Δ O A B = \sqrt{(a^{2} + b^{2}) (r^{2} + a^{2} + b^{2})}

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Solution

The correct options are
A Area of $Δ P A B = \frac{(r^{2} - a^{2} - b^{2})^{\frac{3}{2}}}{\sqrt{a^{2} + b^{2}}}$
D Area of $Δ O A B = \sqrt{(a^{2} + b^{2}) (r^{2} - a^{2} - b^{2})}$
Equation of $A B$ is $a x + b y = a^{2} + b^{2}$
Let the tangents at $A$ and $B$ to the circle intersect at $P (h . k) .$ Then $A B$ is the chord of contact of the tangents drawn from $P (h . k)$ to the circle $x^{2} + y^{2} = r^{2}$ and its equation is
$h x + k y = r^{2} (2)$
As $(1)$ and $(2)$ represent the same line. we have
$\frac{h}{a} = \frac{k}{b} = \frac{r^{2}}{a^{2} + b^{2}} \Rightarrow h = \frac{a r^{2}}{a^{2} + b^{2}}$ and $k = \frac{b r^{2}}{a^{2} + b^{2}}$
Now, the area of the triangle $P A B$ is $P C \times A C$ . where $(P C)^{2} = (h - a)^{2} + (k - b)^{2}$
$= {(\frac{a r^{2}}{a^{2} + b^{2}} - a)}^{2} + {(\frac{b r^{2}}{a^{2} + b^{2}} - b)}^{2}$
$= \frac{(r^{2} - a^{2} - b^{2})^{2} (a^{2} + b^{2})}{(a^{2} + b^{2})^{2}} = \frac{(r^{2} - a^{2} - b^{2})^{2}}{a^{2} + b^{2}}$ and
$(A C)^{2} = (O A^{2}) - (O C^{2}) = r^{2} - (a^{2} + b^{2}) = r^{2} - a^{2} - b^{2}$
Area
of $Δ P A B = \frac{r^{2} - a^{2} - b^{2}}{\sqrt{a^{2} + b^{2}}}, \sqrt{r^{2} - a^{2} - b^{2}} = \frac{(r^{2} - a^{2} - b^{2})^{\frac{3}{2}}}{\sqrt{a^{2} + b^{2}}}$
Area of $Δ O A B = O C \times A C$
$= \sqrt{a^{2} + b^{2}} \sqrt{r^{2} - a^{2} - b^{2}}$

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