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? 
 A: You’re given the slope of one of the tangent lines, so you’ve got its equation: $$x-y\sqrt2=\sqrt2.\tag1$$ Its axis intercepts are $\sqrt2$ and $-1$, respectively. The one-parameter family of circles that pass through these two points have equations $$x(x-\sqrt2)+y(y+1) + \lambda(x-y\sqrt2-\sqrt2)=0.\tag2$$ The first two terms represent a circle with diameter given by the above intercepts. The other two intersections of this circle with the coordinate axes work out to be $(-\lambda,0)$ and $(0,\lambda\sqrt2)$. The line through these two points has an equation of the form $$x\sqrt2-y+\lambda\sqrt2=0.\tag3$$ Even without knowing $\lambda$, you can extract its slope. Plugging the two known slopes into your generic equation of the tangent generates a system of two equations in $a$ and $b$ that you can solve.  
To find the circle through the intercepts, you can substitute $x=2\sqrt2$ and $y=1$ into equation (3), solve for $\lambda$, and plug that into equation (2).  
The end result is illustrated below:

N.B.: This solution assumes that there are four axis intercepts. There’s another solution with a vertical tangent, so that there are only three intercepts.
A: 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,
$$(x-\frac5{2\sqrt2})^2+(y+2)^2=\frac{33}8$$
A: Tangents can intersect the axes at four concyclic points (thanks to amd for correcting me), but they could also intersect at only THREE points: this is not explicitly ruled out by the text of the problem. 
As no tangent can pass through $(0,0)$, this is only possible if a tangent is parallel to $y$-axis, i.e. it has equation $x=2\sqrt2$. Hence $a=2\sqrt2$ and the equations of both tangents are known. one can then easily find that $b=\sqrt3$ all the other requested things.
