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:
Show Hint
- \( (x + \sqrt{2})^2 + y^2 = 3 - \sqrt{2} \)
- \( (x + \sqrt{2})^2 + y^2 = 1 - 2\sqrt{2} \)
- \( (x - \sqrt{2})^2 + y^2 = 2(1 - \sqrt{2}) \)
- \( (x - \sqrt{2})^2 + y^2 = 3 - 2\sqrt{2} \)
The Correct Option is D
Solution and Explanation
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: \[ (x - \sqrt{2})^2 + y^2 = 3 - 2\sqrt{2}. \]
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