Question

If the circumcircle of the triangle formed by the lines ax + by + c = 0 , bx + cy + a = 0 , cx + ay + b = 0 passes through origin prove that
$\left|\begin{array}{ccc}{b}^2& c^2& a^2\\ ab& bc& ca\\ ac& ab& bc\end{array}\right|$ = $\left|\begin{array}{ccc}{b}^2& c^2& a^2\\ bc& ca& ab\\ ac& ab& bc\end{array}\right|$

Answer 1

As.in other parts the equation of the curve which passes through the vertices of the triangle formed by the given lines is
(bx + cy + a) (cx + ay + b) .+ $\lambda$ (cx + ay + b)
(ax + by + c) + $\mu$ (ax + by + c) (bx + cy + a) = 0 ..(1)
If it represents circumcircle of the triangle then coeff. of $x^2$ = coeff. of $y^2$ and coeff of xy = 0
bc.+ $\lambda$ ca + $\mu$ ab = ca + $\lambda$
ab + $\mu$ bc or c(a - b) + $\lambda$a (b - c) + $\mu$ b (c - a) .(2)
and $(c^2+ab)+\lambda (a^2+bc)+\mu (b^2+ca)$ = 0 ...(3)
If (i) passes through (0, 0), then
ab + $\lambda$bc + $\mu$ca = 0 ..(4)
Eliminating $\lambda$, $\mu$ from (2),(3) and (4),we get
$\left|\begin{array}{ccc}ca-bc& ab-ca& bc-ab\\ c^2+ab& a^2+bc& b^2+ca\\ ab& bc& ca\end{array}\right|=0$
Apply $R_2-R_3$
$\left|\begin{array}{ccc}ac-bc& ab-ca& bc-ab\\ c^2& a^2& b^2\\ ab& bc& ca\end{array}\right|=0$

$\left|\begin{array}{ccc}ac& ab& bc\\ c^2& a^2& b^2\\ ab& bc& ca\end{array}\right|=\left|\begin{array}{ccc}bc& ca& ab\\ c^2& a^2& b^2\\ ab& bc& ca\end{array}\right|$
Now inter change $R_1$ and $R_2$ and then make $C_3$ cross over two columns to get the required form.