Question

The equation of the locus of the middle point of a chord of the circle x² + y² = 2(x + y) such that the pair of lines joining the origin to the point of intersection of the chord and the circle are equally inclined to the x-axis is-

• A. x + y = 2
• B. x – y = 2
• C. 2x – y = 1
• D. None

Answer 1

Answer: (A)

x + y = 2

Answer 2

Solving y = mx and x² + y² – 2x – 2y = 0, we get

x² + m²x² – 2x – 2mx = 0

Ž x = 0, $\frac{2(1 + m)}{1 + m^{2}}$

Similarly, solving y = –mx and the equation of the circle, we get

x = 0, $\frac{2(1–m)}{1 + m^{2}}$

\ A = $\left( \frac{2(1 + m)}{1 + m^{2}},\frac{2m(1 + m)}{1 + m^{2}} \right)$

and B = $\left( \frac{2(1–m)}{1 + m^{2}},\frac{- 2m(1 - m)}{1 + m^{2}} \right)$

Let the middle point of AB be (a, b). Then

a = $\frac{1}{2}$ $\left\{ \frac{2(1 + m)}{1 + m^{2}} + \frac{2(1 - m)}{1 + m^{2}} \right\}$ and

b = $\frac{1}{2}$ $\left\{ \frac{2m(1 + m)}{1 + m^{2}} + \frac{- 2m(1 - m)}{1 + m^{2}} \right\}$

\ a = $\frac{2}{1 + m^{2}}$, b = $\frac{2m^{2}}{1 + m^{2}}$. Eliminating m from these,

a + b = 2.