Question

If the tangents drawn at the points O(0, 0) and P(1 + √5, 2) on the circle x² + y² – 2x – 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to

• A. \(\frac{3+\sqrt5}{2}\)
• B. \(\frac{4+2\sqrt5}{2}\)
• C. \(\frac{5+3\sqrt5}{2}\)
• D. \(\frac{7+3\sqrt5} {2}\)

Answer 1

The Correct Option is C

The correct answer is (C) : \(\frac{5+3\sqrt5}{2}\)

Fig. Tangent

\(tan 2θ = 2 ⇒ \frac{2tanθ}{1 - tan²θ} = 2\)
\(tan θ = \frac{\sqrt{5}-1}{ 2}\)
( as θ is acute )

\(Area = \frac{1}{2} L²\)\(sin 2θ = \frac{1}{2} . \frac{5}{tan²θ} . 2sinθcosθ\)
\(= \frac{5sinθcosθ}{sin²θ} . cos²θ\)
= 5cotθ.cos²θ
\(= 5 . \frac{2}{\sqrt5-1} . \frac{1}{1 + ( \frac{\sqrt5-1}{2})²}\)
\(= \frac{10}{\sqrt5-1} . \frac{4}{4+6-2\sqrt5}\)
\(= \frac{40}{2\sqrt5 (\sqrt5-1 )²} = \frac{4\sqrt5}{6-2\sqrt5}\)
\(= \frac{4\sqrt5 ( 6+2\sqrt5 )}{16}\)
\(= \frac{\sqrt5 ( 3+\sqrt5)}{2}\)