Question

The line lx + my = 1 intersects the circle x² + y² = a² at points A, B, if AB subtends 45° at the origin, then a² (l² + m²) -

• A. 4 ± 2$\sqrt{2}$
• B. 4 ±2$\sqrt{6}$
• C. 2$\sqrt{6}$
• D. 4 – $\sqrt{6}$

Answer 1

Answer: (A)

4 ± 2$\sqrt{2}$

Answer 2

Homogenise the equation of the circle with the help of the equation of the line

x² + y² = a² (1)² = x² + y² = a² (lx + my)²

 (a²l² – 1) x² + 2lma²xy + (a²m² – 1) y² = 0

\ tan 45⁰ =$\frac{2\sqrt{l^{2}m^{2}a^{4} - (a^{2}l^{2} - 1)(a^{2}m^{2} - 1)}}{a^{2}l^{2} - 1 + a^{2}m^{2} - 1}$

Squaring, we get 1 = $\frac{4(a^{2}l^{2} + a^{2}m^{2} - 1)}{(a^{2}l^{2} + a^{2}m^{2} - 2)^{2}}$

 a⁴ (l² + m²) – 8a² (l² + m²) + 8 = 0

\ l² + m² = $\frac{8a^{2} \pm \sqrt{64a^{4} - 32a^{4}}}{2a^{4}}$

a² (l² + m²) = 4 ฑ 2$\sqrt{2}$.