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 \((x+\sqrt{2})^2+y^2=2\)
- B \((x-\sqrt{2})^2+(y-\sqrt{3})^2=2\)
- C \((x-\sqrt{2})^2+y^2=(\sqrt{2}-1)^2\)
- D \(x^2+(y-\sqrt{2})^2=(\sqrt{2}+1)^2\)
Answer & Solution
Correct Answer
(C) \((x-\sqrt{2})^2+y^2=(\sqrt{2}-1)^2\)
Step-by-step Solution
Detailed explanation
Let the center of the circle be \((h, k)\) and radius be \(r\). The lines are \(x+y-2=0\) and \(x-y-2=0\). The center must be equidistant from these lines, so \(|h+k-2| = |h-k-2|\). This implies \(k=0\) or \(h=2\). Testing options: Only options A \((-\sqrt{2}, 0)\) and C…
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