value of $\tan B\tan C$ in a triangle 
In a triangle $ABC,$ equation of side $BC$ is $2x-y=3$ and circumcenter and orthocenter of triangle are $(2,4)$ and $(1,2)$ respectively, Then $\tan B\tan C=$

what i try
In a triangle, Let $BC$ be base side and $H(1,2)$ be orthocenter of triangle and $O(2,4)$ be orthocenter of triangle.
and we know that Image of orthocenter $H(1,2)$ of triangle $ABC$ with respect to side $BC$ lie on circumference of  Circumcirlce of $\triangle ABC$
for image point coordinate
using formula
$\displaystyle \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=-\frac{(ax_{1}+by_{1}+c)}{a^2+b^2}$
above image of point $P(x_{1},y_{1})$ w. r to line $ax+by+c=0$ 
$$\frac{x-1}{2}=\frac{y-2}{-1}=-\frac{2(1)-2-3}{2^2+1^2}$$
Coordinate of any point Image point which lie on circumcircle is $$(x,y)=\bigg(\frac{11}{5},\frac{7}{5}\bigg)$$
so circumradius of circle is $$\sqrt{\bigg(2-\frac{11}{5}\bigg)^2+\bigg(4-\frac{7}{5}\bigg)^2}=\frac{\sqrt{170}}{5}$$
How do i solve it Help me please 

If $K$ is the intersection of altitude $AH$ with $BC$, and $H'$ the reflection of $H$ about $BC$, then:
$$
\tan B\cdot\tan C={AK\over BK}\cdot{AK\over CK}={AK^2\over AK\cdot KH'}=
{AK\over KH'}.
$$
