AP EAMCET · Maths · Circle
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 \((x-\sqrt{2})^2+y^2-3 = 2 \sqrt{2}\)
Answer & Solution
Correct Answer
(D) \((x-\sqrt{2})^2+y^2-3 = 2 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
Let \(r\) be the radius of required circle.…
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