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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 
$$\begin{vmatrix} b^2 & c^2  & a^2  \\ ab  & bc  & ca \\ ac  & ab  & bc  \end{vmatrix}$$ = $$\begin{vmatrix} b^2 & c^2 & a^2 \\  bc & ca  & ab  \\ ac  & ab & bc  \end{vmatrix}$$


Solution

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
$$\begin{vmatrix} ca \, - \, bc & ab \, - \, ca & bc \, - \, ab \\ c^2 \, + \, ab & a^2 \, + \, bc & b^2 \, + \, ca \\ ab & bc & ca \end{vmatrix} \, = \, 0$$
Apply $$R_2 \, - R_3$$ 
$$\begin{vmatrix} ac \, - \, bc & ab \, - \, ca & bc \, - \, ab \\ c^2 & a^2 & b^2 \\ ab & bc & ca \end{vmatrix} \, = \, 0$$
$$\,$$
$$\begin{vmatrix} ac & ab & bc \\ c^2 & a^2 & b^2 \\ ab & bc & ca \end{vmatrix} \, = \, \begin{vmatrix} bc & ca & ab \\ c^2 & a^2 & b^2 \\ ab & bc & ca \end{vmatrix}$$
Now inter change $$R_1$$ and $$R_2$$ and then make $$C_3$$ cross over two columns to get the required form.

Mathematics

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