Question

The ratio of the areas of the greatest and the smallest circles touching \((x \pm 1)^2 + (y \pm 1)^2 = 1\) is (The equation \((x \pm 1)^2 + (y \pm 1)^2 = 1\) represents four circles with radius 1, centered at (1,1), (1,-1), (-1,1), (-1,-1).)

• A. \( \frac{\sqrt{3}+1}{\sqrt{3}-1} \)
• B. \( \frac{3+\sqrt{2}}{3-\sqrt{2}} \)
• C. \( \frac{3+2\sqrt{2}}{3-2\sqrt{2}} \)
• D. 4

Hint:

The equations \((x \pm c_x)^2 + (y \pm c_y)^2 = r^2\) describe circles.
For circles touching externally/internally, the distance between centers relates to the sum/difference of radii.
The smallest circle touching four given circles externally is like an incircle to the shape formed by them.
The greatest circle touching four given circles (and enclosing them) is like a circumcircle.
Area of a circle = \(\pi r^2\).

Answer 1

The Correct Option is C

The four given circles are: C1: \((x-1)^2+(y-1)^2=1\), center (1,1), radius r=1. C2: \((x-1)^2+(y+1)^2=1\), center (1,-1), radius r=1. C3: \((x+1)^2+(y-1)^2=1\), center (-1,1), radius r=1. C4: \((x+1)^2+(y+1)^2=1\), center (-1,-1), radius r=1. These four circles form a square arrangement. They touch the x and y axes. The smallest circle touching all four of these circles externally will be centered at the origin (0,0). Let its radius be \(r_{small}\). The distance from origin to center (1,1) is \(\sqrt{1^2+1^2} = \sqrt{2}\). This distance is \(r_{small} + r = r_{small} + 1\). So, \(r_{small}+1 = \sqrt{2} \Rightarrow r_{small} = \sqrt{2}-1\). Area of smallest circle \(A_{small} = \pi r_{small}^2 = \pi (\sqrt{2}-1)^2 = \pi (2 - 2\sqrt{2} + 1) = \pi (3 - 2\sqrt{2})\). The greatest circle touching all four of these circles (likely internally, or an enclosing circle) will also be centered at the origin (0,0). Let its radius be \(r_{greatest}\). This circle will enclose the four given circles and touch them. The distance from origin to the furthest point of any of the circles C1, C2, C3, C4. Consider C1, center (1,1), radius 1. Furthest point from origin on C1 is along the line connecting origin to (1,1), and is \((\sqrt{2})+1\) away from origin. So, \(r_{greatest} = \text{distance from origin to center of C1} + \text{radius of C1} = \sqrt{2} + 1\). Area of greatest circle \(A_{greatest} = \pi r_{greatest}^2 = \pi (\sqrt{2}+1)^2 = \pi (2 + 2\sqrt{2} + 1) = \pi (3 + 2\sqrt{2})\). Ratio of areas \( \frac{A_{greatest}}{A_{small}} = \frac{\pi (3 + 2\sqrt{2})}{\pi (3 - 2\sqrt{2})} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}} \). This matches option (c). To rationalize (optional for matching): \( \frac{3+2\sqrt{2}}{3-2\sqrt{2}} \times \frac{3+2\sqrt{2}}{3+2\sqrt{2}} = \frac{(3+2\sqrt{2})^2}{3^2 - (2\sqrt{2})^2} = \frac{9 + 8 + 12\sqrt{2}}{9 - 8} = \frac{17+12\sqrt{2}}{1} = 17+12\sqrt{2} \). The option is given in unrationalized form. \[ \boxed{\frac{3+2\sqrt{2}}{3-2\sqrt{2}}} \]