<s>[INST] Pair of tangents from point $(2\sqrt2,1)$ to hyperbola $\frac{x^2}{a^2} -\frac{y^2}{b^2} = 1$ 
From a point $(2\sqrt2,1)$ a pair of tangents are drawn to $$\frac{x^2}{a^2} -\frac{y^2}{b^2} = 1$$ which intersect the coordinate axes in concyclic points. If one of the tangents is inclined at an angle of $\arctan\frac{1}{\sqrt{2}}$ with the transverse axis of the hyperbola, then find the equation of the hyperbola and also the circle formed using the concyclic points.

My Attempt
A tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with slope $m$ is given by 
$y=mx±\sqrt{a^2m^2-b^2}$
Plugging $(2\sqrt2,1)$ in this equation, I get
$m^2(8-a^2)+m(-4\sqrt2)+(1+b^2)=0$
This equation gives two values of $m$ 
$m_1=\frac{1}{\sqrt2}$ and $m_2$
$m_1+m_2=\Large\frac{4\sqrt2}{8-a^2}$
And
$m_1m_2=\Large\frac{1+b^2}{8-a^2}$
How do I proceed further? I know we have to use the fact that the points at which the tangents intersect the axes are concyclic. How do I apply this and get the required result or is there another easy way to do this? 
 [/INST] Given the point $(2\sqrt2,1)$ and the slopes $\frac1{\sqrt2}$, $m$, the equations of the two tangent lines are 
$$ y-1 =\frac1{\sqrt2}( x-2\sqrt2), \>\>\>\>\>y-1 = m(x-2\sqrt2)$$
which intersect the axes at $A(\sqrt2,0)$, $B(0,-1)$ and $C(2\sqrt2-\frac1m, 0)$, $D(0, 1-2\sqrt2m)$, respectively.  Given that $A$, $B$, $C$ abd $D$ are concyclic, we have $\angle ACB = \angle ADB=\theta$, i.e.
$$\tan\theta=\frac {BO}{CO}=\frac {AO}{DO}
\implies \frac1{2\sqrt2-\frac1m}=\frac{\sqrt2}{2\sqrt2m-1}$$
which leads to $m=\sqrt2$. The tangent line equations to the hyperbola  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by 
$$(y-m x)^2=a^2m^2-b^2$$
Substitute the point $(2\sqrt2,1)$ and the slopes $m=\frac1{\sqrt2},\>\sqrt2$
 into the equations to get
$$2a^2-b^2=9,\>\>\>\>\>\frac12a^2 -b^2 = 1$$
Solve to obtain $a^2=\frac{16}{3}$, $b^2=\frac53$ and the equation of the hyperbola
$$\frac{3x^2}{16}-\frac{3y^2}{5}=1$$
From the known axis intersections, the cyclic circle is obtain as,
$$(x-\frac5{2\sqrt2})^2+(y+2)^2=\frac{33}8$$
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