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Triangle formed by the lines \mathrm{x}+\mathrm{y}=0, \mathrm{x}-\mathrm{y}=0$ and $l \mathrm{x}+\mathrm{my}=1.If \mathrm{l} 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{x} 2-\mathrm{y} 2) 2=\mathrm{x} 2+\mathrm{y} 2


Option: 2

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


Option: 3

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


Option: 4

(\mathrm{x} 2-\mathrm{y} 2) 2=(\mathrm{x} 2+\mathrm{y} 2) 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}} \quad \ldots .(1) \text { and } l 2+\mathrm{m} 2=1

\mathrm{and \: \mathrm{k}=-\frac{\mathrm{m}}{l^{2}-\mathrm{m}^{2}}\: \ldots(2)}

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

locus is \mathrm{\mathrm{x} 2+\mathrm{y} 2=(\mathrm{x} 2-\mathrm{y} 2) 2}

 

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vinayak

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