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

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(A) $1.$ $A$ tangent to $\frac{x^2}{9}-\frac{y^2}{4}=1$ is $y=mx+\sqrt{9m^2-4}$,where $m>0$.This line is also tangent to the circle $x^2+y^2-8x=0$,whi

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$(B, A)$

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(A) $1.$ $A$ tangent to $\frac{x^2}{9}-\frac{y^2}{4}=1$ is $y=mx+\sqrt{9m^2-4}$,where $m>0$.This line is also tangent to the circle $x^2+y^2-8x=0$,which has center $(4, 0)$ and radius $r=4$.The perpendicular distance from the center $(4, 0)$ to the line $mx-y+\sqrt{9m^2-4}=0$ must equal the radius $4$.$\frac{|4m-0+\sqrt{9m^2-4}|}{\sqrt{m^2+1}}=4 \Rightarrow |4m+\sqrt{9m^2-4}|=4\sqrt{m^2+1}$.Squaring both sides: $16m^2 + 9m^2 - 4 + 8m\sqrt{9m^2-4} = 16(m^2+1) = 16m^2+16$.$8m\sqrt{9m^2-4} = 20-9m^2$.Squaring again: $64m^2(9m^2-4) = (20-9m^2)^2 \Rightarrow 576m^4 - 256m^2 = 400 - 360m^2 + 81m^4$.$495m^4 + 104m^2 - 400 = 0$. Solving for $m^2$,we get $m^2 = 4/5$,so $m = 2/\sqrt{5}$.Substituting $m$ back,the tangent is $y = \frac{2}{\sqrt{5}}x + \sqrt{9(\frac{4}{5})-4} = \frac{2}{\sqrt{5}}x + \sqrt{\frac{16}{5}} = \frac{2}{\sqrt{5}}x + \frac{4}{\sqrt{5}}$.Thus,$2x-\sqrt{5}y+4=0$,which is option $(B)$.$2.$ $A$ point on the hyperbola is $(3\sec	heta, 2	an	heta)$.Substituting into the circle equation: $(3\sec	heta)^2 + (2	an	heta)^2 - 8(3\sec	heta) = 0$.$9\sec^2	heta + 4(\sec^2	heta-1) - 24\sec	heta = 0 \Rightarrow 13\sec^2	heta - 24\sec	heta - 4 = 0$.$(13\sec	heta+2)(\sec	heta-2) = 0$. Since $\sec	heta=2$,$	an^2	heta = 2^2-1=3$,so $	an	heta = \pm\sqrt{3}$.The points are $A(6, 2\sqrt{3})$ and $B(6, -2\sqrt{3})$.The circle with diameter $AB$ has center $(6, 0)$ and radius $2\sqrt{3}$.Equation: $(x-6)^2 + y^2 = (2\sqrt{3})^2 \Rightarrow x^2-12x+36+y^2=12 \Rightarrow x^2+y^2-12x+24=0$,which is option $(A)$.

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