Is a < 2^(1/2) (1) Point (a,0) is inside the circle x^2 + y^2 =3

Solution

Step 1: Analyse Question Stem

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

• The given circle has centre at (0,0) and has a radius of \(\sqrt {3}\)
• If point (a, 0) lies inside the given circle then the distance between the center of the circle and the point must be less than the radius of the circle.

o So, \(\sqrt{(a -0)^2 + (0 – 0)^2} < \sqrt {3}\)
\(⟹ a^2 < 3\)
\(⟹ a^2 - 3 < 0\)
\(⟹ - \sqrt{3} < a < \sqrt{3}\)
• So, multiple values are possible for a. For example, a can be 1 in that case \(a < \sqrt{2} \)or a can 1.6 in that case \(a > \sqrt{2}\)

o Therefore, we cannot be sure if \(a < \sqrt{2}\)

• Since, point (a,1) lies inside the circle \(x^2+y^2=3\)

This means, \(\sqrt{(a- 0)^2 + (1-0)^2} < \sqrt {3}\)
\(⟹ a^2 + 1 < 3\)
\(⟹ a^2 < 2\)
\(⟹ a^2 - 2 < 0\)
\(⟹ - \sqrt {2}< a < \sqrt {2}\)