Question

The radius of a circle touching all the four circles (x ± λ)² + (y ± λ)² = λ² is

• A. 2√2λ
• B. (√2 - 1)λ
• C. (2 + √2)λ
• D. (2- √2)λ

Answer 1

The Correct Option is B

To find the radius \(r\) of a circle that touches the four circles defined by \((x \pm \lambda)^2 + (y \pm \lambda)^2 = \lambda^2\), we proceed as follows:

1. Equations of the Four Circles:
The given equations are \((x \pm \lambda)^2 + (y \pm \lambda)^2 = \lambda^2\). Expand:

\( (x^2 \pm 2\lambda x + \lambda^2) + (y^2 \pm 2\lambda y + \lambda^2) = \lambda^2 \)
\( x^2 + y^2 \pm 2\lambda x \pm 2\lambda y + \lambda^2 = 0 \)
The four combinations yield centers at \((\lambda, \lambda)\), \((\lambda, -\lambda)\), \((-\lambda, \lambda)\), and \((-\lambda, -\lambda)\), each with radius \(\lambda\).

2. Center of the Touching Circle:
Assume the circle that touches all four is centered at the origin \((0, 0)\) with radius \(r\). This is reasonable due to the symmetry of the four centers around the origin.

3. Distance to Each Center:
Calculate the distance from the origin \((0, 0)\) to each circle’s center:

\( \sqrt{(\pm \lambda)^2 + (\pm \lambda)^2} = \sqrt{2\lambda^2} = \sqrt{2}\lambda \)

4. External Tangency Condition:
For the circle to touch each of the four circles externally, the distance between centers equals the sum of their radii:

\( \sqrt{2}\lambda = r + \lambda \)
Solve:

\( r = \sqrt{2}\lambda - \lambda = (\sqrt{2} - 1)\lambda \)
Since \(r > 0\), this solution is valid.

5. Considering Internal Tangency:
For internal tangency, the distance equals the absolute difference of the radii:

\( \sqrt{2}\lambda = |r - \lambda| \)
Solve:

\( r - \lambda = \sqrt{2}\lambda \implies r = (1 + \sqrt{2})\lambda \quad \text{(valid, as } r > 0\text{)} \)
\( r - \lambda = -\sqrt{2}\lambda \implies r = (1 - \sqrt{2})\lambda \quad \text{(invalid, as } r < 0\text{)} \)
However, the external tangency solution \(r = (\sqrt{2} - 1)\lambda\) is specified.

Final Answer:
The radius of the circle that touches all four circles externally is \((\sqrt{2} - 1)\lambda\).