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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 $1, 2$
$(B,A)$
$(B,D)$
$(B,C)$
$(A,D)$
Solution
$1.$ A tangent to $\frac{x^2}{9}-\frac{y^2}{4}=1$ is $y=m x+\sqrt{9 m^2-4}, m>0$
It is tangent to $x^2+y^2-8 x=0$
$\therefore \frac{4 \mathrm{~m}+\sqrt{9 \mathrm{~m}^2-4}}{\sqrt{1+\mathrm{m}^2}}=4$
$\Rightarrow 495 \mathrm{~m}^4+104 \mathrm{~m}^2-400=0$
$\Rightarrow \mathrm{m}^2=\frac{4}{5}$ or $\mathrm{m}=\frac{2}{\sqrt{5}}$
$\therefore$ the tangent is $\mathrm{y}=\frac{2}{\sqrt{5}} \mathrm{~m}+\frac{4}{\sqrt{5}}$
$\Rightarrow 2 \mathrm{x}-\sqrt{5} \mathrm{y}+4=0$.
$2.$ A point on hyperbola is $(3 \sec \theta, 2 \tan \theta)$
It lies on the circle, so $9 \sec ^2 \theta+4 \tan ^2 \theta-24 \sec \theta=0$
$ \Rightarrow 13 \sec ^2 \theta-24 \sec \theta-4=0 \Rightarrow \sec \theta=2,-\frac{2}{13} $
$ \therefore \sec \theta=2 \Rightarrow \tan \theta=\sqrt{3} .$
The point of intersection are $\mathrm{A}(6,2 \sqrt{3})$ and $\mathrm{B}(6,-2 \sqrt{3})$
$\therefore$ The circle with $\mathrm{AB}$ as diameter is
$(x-6)^2+y^2=(2 \sqrt{3})^2 \Rightarrow x^2+y^2-12 x+24=0 \text {. }$
