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  • The circumcircle of the triangle formed by the lines  <img alt="\mathrm{a x+b y+c=0, b x+c y+a=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Ba%20x&plus;b%20y&plu

The circumcircle of the triangle formed by the lines  \mathrm{a x+b y+c=0, b x+c y+a=0} and \mathrm{cx}+\mathrm{ay}+\mathrm{b}=0 passes through the origin if  (\mathrm{k}-2)\left(\mathrm{b}^2+\mathrm{c}^2\right)\left(\mathrm{c}^2+\mathrm{a}^2\right)\left(\mathrm{a}^2+\mathrm{b}^2\right)=\mathrm{abc}(\mathrm{b}+\mathrm{c})(\mathrm{c}+\mathrm{a})(\mathrm{a}+\mathrm{b})where \mathrm{k}=

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Equation of conic is
\mathrm{(b x+c y+a)(c x+a y+b)+\lambda(c x+a y+b)(a x+b y+c)+\mu(a x+b y+c)(b x+c y+a)=0 }
where \mathrm{\lambda \, \, and \, \, \mu } are constants.
Equation (1) represents a circle if the coefficient of  \mathrm{x^2 and\, y^2 } are equal and the coefficient of x y is zero such that
\mathrm{\mathrm{bc}+\lambda \mathrm{ca}+\mu \mathrm{ab}=\mathrm{ca}+\lambda \mathrm{ab}+\mu \mathrm{bc} }
or
\mathrm{(\mathrm{a}-\mathrm{b}) \mathrm{c}+\lambda(\mathrm{b}-\mathrm{c}) \mathrm{a}+\mu(\mathrm{c}-\mathrm{a}) \mathrm{b}=0 }...............(2)
and
\mathrm{\left(c^2+a b\right)+\lambda\left(a^2+b c\right)+\mu\left(b^2+a c\right)=0 }...............(3)
Solving (2) and (3) by cross multiplication rule, we get
\mathrm{\therefore \quad \lambda=\frac{\left(a^2-b c\right)\left(b^2+c^2\right)}{\left(c^2-a b\right)\left(a^2+b^2\right)} \quad \text { and } \mu=\frac{\left(b^2-a c\right)\left(c^2+a^2\right)}{\left(c^2-a b\right)\left(a^2+b^2\right)} }.......(4)
and given, (1) passes through the origin then \mathrm{a b+b c \lambda+c a \mu=0 } ..........(5)
from (4) and (5) we get
\mathrm{\begin{array}{ll} \Rightarrow \quad & a b c^2\left(a^2+b^2\right)+a^2 b c\left(b^2+c^2\right)+b^2 c a\left(c^2+a^2\right) \\ & =a^2 b^2\left(a^2+b^2\right)+b^2 c^2\left(b^2+c^2\right)+c^2 a^2\left(c^2+a^2\right) \\ \Rightarrow & a b c\left\{c\left(a^2+b^2\right)+a\left(b^2+c^2\right)+b\left(c^2+a^2\right)\right\} \end{array} }
 

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Rishabh

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