Tangents are drawn from any point on the hyperbola $\frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1$to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=9.$ If the locus of the mid-point of the chord of contact is $\mathrm{a}{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2 = {\mathrm{bx}}^2 – {\mathrm{cy}}^2,$ then the value of ${\mathrm{a}}^2 +{\mathrm{b}}^2+{\mathrm{c}}^2 =$

$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1 \\ \mathrm{let} \mathrm{P}=(3\mathrm{sec\theta }, 2\mathrm{tan\theta }) \mathrm{be} \mathrm{any} \mathrm{point} \mathrm{on} \mathrm{given} \mathrm{hyperbola} \\ \mathrm{Now} \mathrm{chord} \mathrm{of} \mathrm{contact} \mathrm{of} \text{'}\mathrm{P}\text{'} \mathrm{w}.\mathrm{r}. \mathrm{to} \mathrm{the} \mathrm{circle} {\mathrm{x}}^2+{\mathrm{y}}^2=9 \mathrm{is} \mathrm{x}\left(3\mathrm{sec\theta }\right)+\mathrm{y}\left(2\mathrm{tan\theta }\right)=9 \\ \Rightarrow 3\mathrm{x} \mathrm{sec\theta }+2\mathrm{y} \mathrm{tan\theta }=9 — ① \\ \mathrm{let} ({\mathrm{x}}_1, {\mathrm{y}}_1) \mathrm{be} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{chord} \mathrm{of} \mathrm{contact} \mathrm{then} \mathrm{equation} \mathrm{of} \mathrm{chord} \mathrm{in} \mathrm{mid} \mathrm{point} \mathrm{form} \mathrm{is} \\ {\mathrm{xx}}_1+{\mathrm{yy}}_1={{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2 — ② \\ \mathrm{since} ① \& ② \mathrm{represents} \mathrm{same} \mathrm{line} \mathrm{then} \\ \frac{3\mathrm{sec\theta }}{{\mathrm{x}}_1}=\frac{2\mathrm{tan\theta }}{{\mathrm{y}}_1}=\frac{9}{{{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2} \\ \Rightarrow \mathrm{sec\theta }=\frac{9{\mathrm{x}}_1}{3({{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2)}, \tan \theta =\frac{9{\mathrm{y}}_1}{2({{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2)} \end{gathered}$$

$$\begin{gathered} \mathrm{Since} {\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1 \\ \Rightarrow \frac{81{{\mathrm{x}}_1}^2}{9({{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2)}-\frac{81{{\mathrm{y}}_{14}}^2}{4({{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2)}=1 \\ \Rightarrow \frac{81}{({{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2)}\left(\frac{{{\mathrm{x}}_1}^2}{9}-\frac{{{\mathrm{y}}_1}^2}{4}\right)=1 \\ \therefore \mathrm{locus} \mathrm{is} \frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=\frac{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}{81} \\ \Rightarrow 4{\mathrm{x}}^2-9{\mathrm{y}}^2=\frac{36{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}{81} \\ \Rightarrow 4{\mathrm{x}}^2-9{\mathrm{y}}^2=\frac{4}{9}({\mathrm{x}}^2+{\mathrm{y}}^2) \Rightarrow 36{\mathrm{x}}^2-81{\mathrm{y}}^2=4{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2 \\ \mathrm{By} \mathrm{comparision} \mathrm{a}=4, \mathrm{b}=36, \mathrm{c}=81 \\ {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2=96+1296+6561 \\ =7873 \end{gathered}$$