JEE Main 2022 · 28 Jun · Shift 1 · Mathematics Q14

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MathematicsSingle correct+4 / -1Question 14 of 90

If the tangents drawn at the points $O(0,0)$ and $P\left(1+\sqrt5,2\right)$ on the circle $x^2+y^2-2x-4y=0$ intersect at the point Q, then the area of the triangle OPQ is equal to

\dfrac{3+\sqrt5}2

\dfrac{4+2\sqrt5}2

\dfrac{5+3\sqrt5}2

\dfrac{7+3\sqrt5}2

Answer and worked solution

Official answer: C

Tangents. At O: $x+2y=0$ . At P: $\sqrt5x=5+\sqrt5$ , i.e. $x=1+\sqrt5$ .

Q $=\left(1+\sqrt5,-\frac{1+\sqrt5}2\right)$ , directly below P.

Area. $\frac12\cdot PQ\cdot(1+\sqrt5)=\frac12\cdot\frac{5+\sqrt5}2\cdot(1+\sqrt5)=\frac{5+3\sqrt5}2$ .

(C)

Topics:Tangents and normals

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JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 14