Question

The equation of a circle which touches the straight lines \[ x + y = 2, \quad x - y = 2 \] and also touches the circle \[ x^2 + y^2 = 1 \] is:

• A. \( (x + \sqrt{2})^2 + y^2 = 3 - \sqrt{2} \)
• B. \( (x + \sqrt{2})^2 + y^2 = 1 - 2\sqrt{2} \)
• C. \( (x - \sqrt{2})^2 + y^2 = 2(1 - \sqrt{2}) \)
• D. \( (x - \sqrt{2})^2 + y^2 = 3 - 2\sqrt{2} \)

Hint:

For a circle to touch two lines and another circle, use perpendicular distance conditions and radius calculations.

Answer 1

The Correct Option is D

Step 1: Understanding the given conditions The given lines \( x+y = 2 \) and \( x-y = 2 \) are perpendicular, forming a square with the given circle.
Step 2: Finding the required circle equation Using the standard form of a circle and solving for the appropriate radius satisfying the tangency conditions, we get: \[ (x - \sqrt{2})^2 + y^2 = 3 - 2\sqrt{2}. \]