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If the lines a1x+b1y+c1=0 and a2x+b2y+c2...

If the lines `a_1x+b_1y+c_1=0` and `a_2x+b_2y+c_2=0` cut the coordinae axes at concyclic points, then prove that `|a_1a_2|=|b_1b_2|`

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Line `a_(1)x+b_(1)y+c_(1)=0` meets he axes at points `P(-(c_(1))/(a_(1)),0)`
and `Q(0,-(c_(1))/(b_(1)))`
Line `a_(2)x+b_(2)y+c_(2)=0` meets the axes at points `R(-(c_(2))/(a_(2)),0)` and `S(0,-(c_(2))/(b_(2)))`
Since points P,Q,R and S are concyclic , we have
`OP xx OR=OQ xx OS` , where O is origin.
`implies |(-(c_(1))/(a_(1)))(-(c_(2))/(a_(2)))|=|(-(c_(1))/(b_(1)))(-(c_(2))/(b_(2)))|`
`implies |a_(1)a_(2)|=|b_(1)b_(2)|`
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