If the line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ is a common tangent to the hyperbola $\frac{{\mathrm{x}}^2}{100}-\frac{{\mathrm{y}}^2}{64}=1$ and the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=36$, then which one of the following is true ?

$$\begin{gathered} Given hyperbola is \frac{x^2}{100}-\frac{y^2}{64}=1 \\ where a^2=100,b^2=64 \\ Given circle x^2+y^2=36 where r^2=36 \end{gathered}$$

General tangent to the hyperbola in slope form is $\mathrm{y}=\mathrm{mx}\pm \sqrt{100{\mathrm{m}}^2-64}$ and the general tangent to the circle in slope form is $y=mx\pm 6\sqrt{1+m^2}$

For common tangent, $36\left(1+{\mathrm{m}}^2\right)=100{\mathrm{m}}^2-64$

$\begin{array}{l}\Rightarrow 100=64{\mathrm{m}}^2\Rightarrow {\mathrm{m}}^2=\frac{100}{64}\\ \therefore {\mathrm{c}}^2=36\left(1+\frac{100}{64}\right)=\frac{164\times 36}{64}=\frac{369}{4}\end{array}$