Question

The equation of a circle which touches the straight lines $x + y = 2$, $x - y = 2$ 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: \[ (x - \sqrt{2})^2 + y^2 = 3 - 2\sqrt{2}. \]

Answer 2

Given:
- The circle touches the lines \( x + y = 2 \) and \( x - y = 2 \).
- The circle also touches the circle \( x^2 + y^2 = 1 \).

Step 1: Since the two lines are perpendicular and intersect at point \( (1, 1) \), the circle touching both lines will have its center on the angle bisector of these lines.

Step 2: The lines can be rewritten as:
\[ x + y - 2 = 0, \quad x - y - 2 = 0 \] The angle bisectors of these lines are:
\[ x = 1 \quad \text{and} \quad y = 1 \] The circle touching both lines will have its center on one of these bisectors. Assume center at \( (h, 0) \) or \( (h, k) \) and find exact position.

Step 3: Let the center be \( (a, 0) \) (since the figure is symmetric in \( y \)).
Distance from center to line \( x + y = 2 \):
\[ d_1 = \frac{|a + 0 - 2|}{\sqrt{1^2 + 1^2}} = \frac{|a - 2|}{\sqrt{2}} \] Distance from center to line \( x - y = 2 \):
\[ d_2 = \frac{|a - 0 - 2|}{\sqrt{1^2 + (-1)^2}} = \frac{|a - 2|}{\sqrt{2}} \] Since circle touches both lines, radius \( r = d_1 = d_2 = \frac{|a - 2|}{\sqrt{2}} \).

Step 4: The circle touches the circle \( x^2 + y^2 = 1 \).
Distance between centers:
\[ \sqrt{(a - 0)^2 + (0 - 0)^2} = |a| \] Sum or difference of radii must equal distance between centers.
Since the circle lies outside \( x^2 + y^2 = 1 \), the external touching condition:
\[ |a| = r + 1 = \frac{|a - 2|}{\sqrt{2}} + 1 \]

Step 5: Solve for \( a \):
Multiply both sides by \( \sqrt{2} \):
\[ \sqrt{2} |a| = |a - 2| + \sqrt{2} \] Consider \( a < 2 \), then \( |a - 2| = 2 - a \), so:
\[ \sqrt{2} a = 2 - a + \sqrt{2} \implies \sqrt{2} a + a = 2 + \sqrt{2} \implies a (\sqrt{2} + 1) = 2 + \sqrt{2} \] \[ a = \frac{2 + \sqrt{2}}{\sqrt{2} + 1} = \sqrt{2} \] (Using rationalization).

Step 6: Radius \( r \) is:
\[ r = \frac{|a - 2|}{\sqrt{2}} = \frac{| \sqrt{2} - 2 |}{\sqrt{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1 \] Square radius:
\[ r^2 = (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} \]

Step 7: Equation of the circle:
\[ (x - \sqrt{2})^2 + y^2 = 3 - 2 \sqrt{2} \]

Therefore, the required equation is:
\[ \boxed{ (x - \sqrt{2})^2 + y^2 = 3 - 2\sqrt{2} } \]