Question

$P,Q,R,S$ are the centres of the four circles each of which is cut by a fixed circle orthogonally. If $t_1^2,t_2^2,t_3^2,t_4^2$ be the squares of the lengths of tangents to the four circles from a point in their plane, then prove that $t_1^2△QRS+t_2^2△RSP+t_3^2△SPQ+t_4^2△PQR=0$.

Answer 1

$P,Q,R,S$ are centres of circles

$x^2+y^2+2g_{r}x+2f_{r}y+c_r=0$,

where $r=1,2,3,4\therefore P$ is $(-g_1,-f_1)$

Now choose any fixed point in the plane of the circles as $(0,0)$, then $t^2=S^{\prime }$

$\therefore t_1^2=c_1,t_2^2=c_2,t_3^2=c_3,t_4^2=c_4$

Let the circle which cuts each of them orthogonally be

$x^2+y^2+2gx+2fy+c=0$

$\therefore 2g{g}_1+2f{f}_1=c+c_1=c+t_1^2$

or $2g{g}_1+2f{f}_1-c-t_1^2=0.....\left(1\right)$

There will be four such relations as $\left(1\right)$ and we have to eliminate the three unknowns $g,f$ and $c$ between these four.

$\therefore \left|\begin{array}{cccc}2g_1& 2f_1& -1& -t_1^2\\ 2g_2& 2f_2& -1& -t_2^2\\ 2g_3& 2f_3& -1& -t_3^2\\ 2g_4& 2f_4& -1& -t_4^2\end{array}\right|=0$

$\left|\begin{array}{cccc}{t}_1^2& g_1& f_1& 1\\ t_2^2& g_2& f_2& 1\\ t_3^2& g_3& f_3& 1\\ t_4^2& g_4& f_4& 1\end{array}\right|=0$

$t_1^2|D_1|-t_2^2|D_2|+t_3^2|D_3|-t_4^2|D_4|=0.....\left(2\right)$

The first determinant $D_1$ is

$\left|\begin{array}{ccc}{g}_2& f_2& 1\\ g_3& f_3& 1\\ g_4& f_4& 1\end{array}\right|=2△QRS$

$\because △QRS=\frac{1}{2}\left|\begin{array}{ccc}-g_2& -f_2& 1\\ -g_3& -f_3& 1\\ -g_4& -f_4& 1\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}{g}_2& f_2& 1\\ g_3& f_3& 1\\ g_4& f_4& 1\end{array}\right|$

Hence from $\left(2\right)$ we prove the required result.