TS EAMCET · Maths · Circle
The area (in sq units) of the triangle formed by the tangent, normal at \((1, \sqrt{3})\) to the circle \(x^2+y^2=4\) and the \(X\)-axis, is
- A \(4 \sqrt{3}\)
- B \(\frac{7}{2} \sqrt{3}\)
- C \(2 \sqrt{3}\)
- D \(\frac{1}{2} \sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
The equations of the tangent and normal to the circle \(x^2+y^2=4\) at \(P(1, \sqrt{3})\) are \( x+\sqrt{3} y=4 \text { and } y=\sqrt{3} x \) The tangent meets \(X\)-axis at \(A(4,0)\). \(\therefore\) Area of \(\triangle O A P=\frac{1}{2} \times 4 \times \sqrt{3}=2 \sqrt{3}\) sq…
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