JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let a triangle \(A B C\) be inscribed in the circle \(x ^{2}-\) \(\sqrt{2}(x+y)+y^{2}=0\) such that \(\angle B A C=\frac{\pi}{2}\). If the length of side \(A B\) is \(\sqrt{2}\), then the area of the \(\triangle ABC\) is equal to
- A \((\sqrt{2}+\sqrt{6}) / 3\)
- B \((\sqrt{6}+\sqrt{3}) / 2\)
- C \((3+\sqrt{3}) / 4\)
- D \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
Radius of given circle is \(1\) \(BC =\) diameter \(=2, AB =\sqrt{2}\) \(AC =\sqrt{ BC ^{2}- AB ^{2}}=\sqrt{2}\) \(\Delta ABC =\frac{1}{2} AB \cdot AC =1\)
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