Question:
Consider four circles $(x ± 1)² + (y ± 1)² = 1$, then the equation of smaller circle touching these four circles is
Consider four circles $(x ± 1)² + (y ± 1)² = 1$, then the equation of smaller circle touching these four circles is
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Consider four circles $(x ± 1) = 1$, then the equation of smaller circle touching these four circles is
Updated On: Apr 15, 2026
- $x^{2}+y^{2}=3-\sqrt{2}$
- $x^{2}+y^{2}=6-3\sqrt{2}$
- $x^{2}+y^{2}=5-2\sqrt{2}$
- $x^{2}+y^{2}=3-2\sqrt{2}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The centres of the four circles are $(\pm 1, \pm 1)$ and their radius is 1.
Step 2: Analysis
The distance from the origin $(0,0)$ to any centre is $\sqrt{1^{2}+1^{2}} = \sqrt{2}$.
Step 3: Evaluation
The distance from the origin to the nearest point of any circle is $\sqrt{2} - 1$. This is the radius '$r$' of the small inner circle.
Step 4: Conclusion
Equation: $x^{2}+y^{2} = r^{2} = (\sqrt{2}-1)^{2} = 2+1-2\sqrt{2} = 3-2\sqrt{2}$.
Final Answer: (d)
The centres of the four circles are $(\pm 1, \pm 1)$ and their radius is 1.
Step 2: Analysis
The distance from the origin $(0,0)$ to any centre is $\sqrt{1^{2}+1^{2}} = \sqrt{2}$.
Step 3: Evaluation
The distance from the origin to the nearest point of any circle is $\sqrt{2} - 1$. This is the radius '$r$' of the small inner circle.
Step 4: Conclusion
Equation: $x^{2}+y^{2} = r^{2} = (\sqrt{2}-1)^{2} = 2+1-2\sqrt{2} = 3-2\sqrt{2}$.
Final Answer: (d)
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