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Question

Mathematics Question on Circle

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

A

3+52\frac{3+\sqrt5}{2}

B

4+252\frac{4+2\sqrt5}{2}

C

5+352\frac{5+3\sqrt5}{2}

D

7+352\frac{7+3\sqrt5} {2}

Answer

5+352\frac{5+3\sqrt5}{2}

Explanation

Solution

The correct answer is (C) : 5+352\frac{5+3\sqrt5}{2}

Fig. Tangent

tan2θ=2⇒2tanθ1−tan2θ=2tan 2θ = 2 ⇒ \frac{2tanθ}{1 - tan²θ} = 2
tanθ=5−12tan θ = \frac{\sqrt{5}-1}{ 2}
( as θ is acute )

Area = \frac{1}{2} L²$$sin 2θ = \frac{1}{2} . \frac{5}{tan²θ} . 2sinθcosθ
=5sinθcosθsin2θ.cos2θ= \frac{5sinθcosθ}{sin²θ} . cos²θ
= 5cotθ.cos²θ
=5.25−1.11+(5−12)2= 5 . \frac{2}{\sqrt5-1} . \frac{1}{1 + ( \frac{\sqrt5-1}{2})²}
=105−1.44+6−25= \frac{10}{\sqrt5-1} . \frac{4}{4+6-2\sqrt5}
=4025(5−1)2=456−25= \frac{40}{2\sqrt5 (\sqrt5-1 )²} = \frac{4\sqrt5}{6-2\sqrt5}
=45(6+25)16= \frac{4\sqrt5 ( 6+2\sqrt5 )}{16}
=5(3+5)2= \frac{\sqrt5 ( 3+\sqrt5)}{2}