<p>Equations of four circles are $(x \pm a)^2 + (y \pm a)^2 = a^2$, then:</p>
<p>(a) The radius of the greatest circle touching all the four circles is $(\sqrt{2} + 1)a$</p>
<p>(b) The radius of the smallest circle touching all the four circles is $(\sqrt{2} - 1)a$</p>
<p>(c) Area of region enclosed by four given circles is $(4 - \pi)a^2$ sq. units</p>
<p>(d) The centres of four circles are the vertices of a square</p>
Step-by-Step Solution
Key Concept: Recognize the symmetric configuration of four circles and use distance formula to find radii of circles tangent to all four.
<p><strong>Analysis:</strong> The four circles have centres at $(a,a)$, $(a,-a)$, $(-a,a)$, $(-a,-a)$, each with radius $a$. These four points form a square with side length $2a$. For the greatest circle touching all four externally, the radius is distance from origin to centre minus radius of small circle: $r_{max} = a\sqrt{2} + a = (\sqrt{2}+1)a$. For the smallest circle touching all four internally, $r_{min} = a\sqrt{2} - a = (\sqrt{2}-1)a$. The centres indeed form a square.</p><p>∴ Answers are (a), (b), (d).</p>
Correct Answer: a, b, d