Question

Triangle formed by the lines x + y = 0, x – y = 0 and

lx + my = 1. If l and m vary subject to the condition

l² + m² = 1, then the locus of its circumcentre is-

• A. (x² – y²)² = x² + y²
• B. x² + y² = 4x²y²
• C. (x² + y²)² = x² – y²
• D. (x² – y²)² = (x² + y²)²

Answer 1

Answer: (A)

(x² – y²)² = x² + y²

Answer 2

D is right angled with right angle at (0, 0) so for other vertices with y = x $\left( \frac{1}{\mathcal{l} + m},\frac{1}{\mathcal{l} + m} \right)$

with y = – x $\left( \frac{1}{\mathcal{l -}m}, - \frac{1}{\mathcal{l -}m} \right)$

Now circumcentre (h, k)

2h = $\frac{1}{\mathcal{l +}m}$+ $\frac{1}{\mathcal{l -}m}$ & 2k =$\frac{1}{\mathcal{l +}m}$– $\frac{1}{\mathcal{l -}m}$

so (h² – k²)² = k² + h²

locus (x² – y²)² = x² + y²