Question

Let a circle be 2x(x – a) + y (2y – b) = 0, a ¹ 0, b ¹ 0. Then the condition on a and b if two chords each bisected by the x-axis, can be drawn to the circle from$\left( a,\frac{b}{2} \right)$, is-

• A. a²> 2b²
• B. a²< 2b²
• C. a² = 2b²
• D. None

Answer 1

Answer: (A)

a²> 2b²

Answer 2

The equation of the circle is

2x(x – a) + y(2y – b) = 0

Ž x² + y² – ax – $\frac{b}{2}$y = 0

Let PQ and PR be two chords drawn from P$\left( a,\frac{b}{2} \right)$such that they are bisected by x-axis.

Let A(t, 0) be the mid-point of PQ. Then, its equation is

tx + 0y – $\frac{a}{2}$ (x + t) – $\frac{b}{4}$ (y + 0) = t² – at

[Using T = S’]

Ž $\left( t - \frac{a}{2} \right)$x – $\frac{b}{4}$y = t² – $\frac{a}{2}$t

This passes through P$\left( a,\frac{b}{2} \right)$. Therefore,

$\left( t - \frac{a}{2} \right)a - \frac{b^{2}}{8}$ = t² – $\frac{a}{2}$t

Ž t² – $\frac{3}{2}$at + $\left( \frac{a^{2}}{2} + \frac{b^{2}}{8} \right)$ = 0

This should give two distinct values of t for points A and B.

\ $\frac{9}{4}$a² – 4 $\left( \frac{a^{2}}{2} + \frac{b^{2}}{8} \right)$ > 0

Ž $\frac{a^{2}}{4} - \frac{b^{2}}{2} > 0$

Ž a² > 2b²

Hence (1) is the correct answer.