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  •  Triangle formed by the lines <img alt="\mathrm{x}+\mathrm{y}=0, \mathrm{x}-\mathrm{y}=0" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%7D&plus;%5Cmathrm%7By%7D%3D0%2C

 Triangle formed by the lines \mathrm{x}+\mathrm{y}=0, \mathrm{x}-\mathrm{y}=0 and l \mathrm{x}+\mathrm{my}=1. If 1 and \mathrm{m} vary subject to the condition \mathrm{l^{2}+\mathrm{m}^{2}=1} then the locus of its circumcentre is 

Option: 1

\mathrm{\left(x^{2}-y^{2}\right)^{2}=x^{2}+y^{2}}


Option: 2

\mathrm{\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)}


Option: 3

\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=4 \mathrm{x}^{2} \mathrm{y}^{2}


Option: 4

\mathrm{\left(x^{2}-y^{2}\right)^{2}=\left(x^{2}+y^{2}\right)^{2}}


Answers (1)

best_answer

 Coordinates of circumcentre \frac{l}{l^{2}-\mathrm{m}^{2}}, \frac{\mathrm{m}}{\mathrm{m}^{2}-l^{2}}
\mathrm{h}=\frac{l}{l^{2}-\mathrm{m}^{2}} \ldots .(1) \text { and } l^{2}+\mathrm{m}^{2}=1
\text{and }\mathrm{k}=-\frac{\mathrm{m}}{l^{2}-\mathrm{m}^{2}}\quad \cdots{(2)}

square and add (1) and (2)

\mathrm{h}^{2}+\mathrm{k}^{2}=\frac{l^{2}+\mathrm{m}^{2}}{\left(l^{2}-\mathrm{m}^{2}\right)^{2}}=\frac{1}{\left(l^{2}-\mathrm{m}^{2}\right)^{2}}

\therefore \quad \frac{1}{\left(l^{2}-\mathrm{m}^{2}\right)^{2}}=\left(\mathrm{h}^{2}-\mathrm{k}^{2}\right)^{2}

\mathrm{locus \: is \, x^{2}+y^{2}=\left(x^{2}-y^{2}\right)^{2}}


 

 

Posted by

Ajit Kumar Dubey

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