The circle x^2+y^2-8 x=0 and hyperbola frac{x | vedclass

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Question

$1.$  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 $x^2+y^2-8 x=0$

$	herefore \frac{4 \

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$(B,A)$

Vedclass · Answer

$1.$  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 $x^2+y^2-8 x=0$

$	herefore \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 $\mathrm{m}=\frac{2}{\sqrt{5}}$

$	herefore$ 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 $(3 \sec 	heta, 2 	an 	heta)$

It lies on the circle, so $9 \sec ^2 	heta+4 	an ^2 	heta-24 \sec 	heta=0$

$ \Rightarrow 13 \sec ^2 	heta-24 \sec 	heta-4=0 \Rightarrow \sec 	heta=2,-\frac{2}{13} $

$ 	herefore \sec 	heta=2 \Rightarrow 	an 	heta=\sqrt{3} .$

The point of intersection are $\mathrm{A}(6,2 \sqrt{3})$ and $\mathrm{B}(6,-2 \sqrt{3})$

$	herefore$ The circle with $\mathrm{AB}$ as diameter is

$(x-6)^2+y^2=(2 \sqrt{3})^2 \Rightarrow x^2+y^2-12 x+24=0 	ext {. }$

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