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?