Tangents are drawn from the point  $(4,3)$  to the circle  $x^2+y^2=9.$  Then the area of the triangle formed by these tangents and the chord of contact is $\frac{64\Delta }{25}$  square units where  $\Delta$  is equal to

The equation of the chord of contact of the two tangents that can be drawn from $P(4,3)$  to the circle $x^2+y^2=9is(S_1=0)$  
$4x+3y-9=0 \mathrm{.......}\left(1\right)$ 
For area of $\Delta PAB,$  we need
$\left|AB\right|and\left|MP\right|$ 
Here, $\left|OP\right|=\sqrt{{(4-0)}^2+{(3-0)}^2}=5$ 
Also, $\left|OM\right|$  = distance of $O(0,0)$  from line (1)
 
 $=\frac{|0+0-9|}{\sqrt{4^2+3^2}}=\frac{9}{5}$
$\therefore \left|AB\right|=2\sqrt{{\left(radius\right)}^2-O{M}^2}$ 
 $=2\sqrt{3^2-{\left(\frac{9}{5}\right)}^2}=2\sqrt{\frac{144}{25}}=\frac{24}{5}$
and  $\left|MP\right|=\left|OP\right|-\left|OM\right|$
$=5-\frac{9}{5}=\frac{16}{5}$ 
Here area  $\Delta PAB=\frac{1}{2}\left|AB\right|\left|MP\right|$
$=\frac{1}{2}\left(\frac{24}{5}\right)\left(\frac{16}{5}\right)=\frac{192}{25}\text{square\hspace{0.17em}units}$ 
$\Rightarrow \frac{64\Delta }{25}=\frac{192}{25}\Rightarrow \Delta =3.$