Question

The locus of the middle points of the chords of the circle

x² + y² = a² which subtend a right angle at the centre, is –

• A. (x² + y²) – a² = 0
• B. 2(x² + y²) – 2a² = 0
• C. 2(x² – y²) + a² = 0
• D. 2(x² + y²) – a² = 0

Answer 1

Answer: (D)

2(x² + y²) – a² = 0

Answer 2

Let (h, k) be the mid-point of a chord AB of the circle

x² + y² = a². Then, the equation of AB is

hx + ky – a² = h² + k² – a²[Using T = S¢]

Ž hx + ky = h² + k² … (1)

The combined equation of OA and OB is

x² + y² = a² $\left( \frac{hx + ky}{h^{2} + k^{2}} \right)^{2}$

Ž (h² + k²)² (x² + y²) – a² (hx + ky)² = 0

OA and OB will be perpendicular if

Coefficient of x² + Coefficient of y² = 0

Ž (h² + k²)² – a²h² + (h² + k²)² – a²k² = 0

Ž 2(h² + k²)² – a²(h² + k²) = 0

Ž 2(h² + k²) – a² = 0.

So, locus of (h, k) is 2(x² + y²) – a² = 0.