The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$

$2.$ Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

$(A)$ $2 x-\sqrt{5} y-20=0$ $(B)$ $2 x-\sqrt{5} y+4=0$

$(C)$ $3 x-4 y+8=0$ $(D)$ $4 x-3 y+4=0$

$2.$ Equation of the circle with $\mathrm{AB}$ as its diameter is

$(A)$ $x^2+y^2-12 x+24=0$ $(B)$ $x^2+y^2+12 x+24=0$

$(C)$ $\mathrm{x}^2+\mathrm{y}^2+24 \mathrm{x}-12=0$ $(D)$ $x^2+y^2-24 x-12=0$

Give hte answer question $1, 2$

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be $\frac{5}{4}$. If the equation of the normal at the point $\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ on the hyperbola is $8 \sqrt{5} x +\beta y =\lambda$, then $\lambda-\beta$ is equal to

The equation of the common tangents to the two hyperbola $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$ and $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$ are-

The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola $\frac{{{x^2}}}{4} – \frac{{{y^2}}}{5} = 1$ , meet $x-$ axis and $y-$ axis at $A$ and $B$ respectively. Then $(OA)^2 – (OB)^2$ , where $O$ is the origin, equals

Product of length of the perpendiculars drawn from foci on any tangent to hyperbola ${x^2} – \frac{{{y^2}}}{4}$ = $1$ is