Question

Three concentric circles of which biggest is x² + y² = 1, have their radii in A.P. If the line y = x + 1 cuts all the circles in real and distinct distance points, then the interval in which the common difference of A.P. will lie, is –

• A. $\left( 0,\left( 1 - \frac{1}{\sqrt{2}} \right) \right)$
• B. $\left( 0,\frac{1}{2} \right)$
• C. (1, 1)
• D. $\left( 0,\frac{1}{2}\left( 1 - \frac{1}{\sqrt{2}} \right) \right)$

Answer 1

Answer: (D)

$\left( 0,\frac{1}{2}\left( 1 - \frac{1}{\sqrt{2}} \right) \right)$

Answer 2

The equation of the biggest circle is

x² + y² = 1²

Clearly, it is centred at O (0, 0) and has radius 1.

Let the radii of the other two circles be 1 – r, 1 – 2r, where

r > 0.

Thus, the equations of the concentric circles are :

x² + y² = 1 … (1)

x² + y² = (1 – r)² … (2)

x² + y² = (1 – 2r)² … (3)

Clearly, y = x + 1 cuts the circle (1) at (1, 0) and (0, 1). This line will cut circles (2) and (3) in real and distinct points if

$\left| \frac{1}{\sqrt{2}} \right|$ < 1 – r and $\left| \frac{1}{\sqrt{2}} \right|$ < 1 – 2r

Ž $\frac{1}{\sqrt{2}}$ < 1 – r and $\frac{1}{\sqrt{2}}$ < 1 – 2r

Ž r < 1 –$\frac{1}{\sqrt{2}}$and r < $\frac { 1 } { 2 }$ $\left( 1 - \frac{1}{\sqrt{2}} \right)$

Ž r < $\frac { 1 } { 2 }$ $\left( 1 - \frac{1}{\sqrt{2}} \right)$

Ž r Ī $\left( 0,\frac{1}{2}\left( 1 - \frac{1}{\sqrt{2}} \right) \right)$ [Q r > 0]

Hence (4) is the correct answer.