The chord of contact of  $(\mathrm{acos}\mathrm{\theta },\mathrm{asin}\mathrm{\theta })\text{ w.r.t }{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{b}}^2\text{ touches }{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{c}}^2\cdot$$\mathrm{a},\mathrm{b},\mathrm{c}$ are the roots of x³ –14x² + 56x + d = 0 then a is

$$\begin{gathered} x\left(a\cos \theta \right)+y\left(a\sin \theta \right)=b^2 is \tan gent to the circle x^2+y^2=c^2 \\ \Rightarrow c=\frac{\left|b^2\right|}{\sqrt{a^2}({\cos }^2\theta +{\sin }^2\theta )}\therefore b^2=ac \end{gathered}$$

Clearly b² = ac → a,b,c are in G.P given a, b, c are roots of x³-14x² +56x+ d=0  then
$\begin{array}{l}a=\frac{b}{r},c=br(a>b>c)\\ S_{3=}\frac{b}{r}\left(b\right)\left(br\right)=b^3=-d\\ \because b is a root of x^3 –14x^2+ 56x + d = 0\\ b^3-14{\left(b\right)}^2+56\left(b\right)+d=0\Rightarrow 56\left(b\right)=14{\left(b\right)}^2\Rightarrow b=4 is middle root\\ \Rightarrow \frac{b}{r}+b+br=14\Rightarrow \frac{4}{r}+4+4r=14\\ \Rightarrow r+\frac{1}{r}=\frac{5}{2}\\ \therefore {\mathrm{x}}^3-14{\mathrm{x}}^2+56\mathrm{x}-64=0\text{ roots are }8,4,2\\ \left(4\text{ middle root }\right)\\ \Rightarrow \therefore \mathrm{a}=8\end{array}$