The tangent and the normal lines at the point $(\sqrt{3}, 1)$ to the circle $x^{2}+y^{2}=4$ and the $x$-axis form a triangle. The area of this triangle (in square units) is:

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Circle

The tangent and the normal lines at the point

(3, 1)

to the circle

x^{2} + y^{2} = 4

and the

x

-axis form a triangle. The area of this triangle (in square units) is:

(A) $\frac{1}{3}$
(B) $\frac{4}{3}$
(C) $\frac{1}{3}$
(D) $\frac{2}{3}$ ✓ Correct

Answer: (D)

Solution

P (3, 1)

on the circle

x^{2} + y^{2} = 4

the radius

O P

has slope

\frac{1}{3}

, so the tangent, being perpendicular to it, has slope

- 3

. Its equation is

y - 1 = - 3 (x - 3)

, that is

3 x + y = 4

, which meets the x-axis at

Q (\frac{4}{3}, 0)

. The normal is the line through the centre and

P

, passing through the origin

O

. The triangle

O P Q

has its base

O Q = \frac{4}{3}

along the x-axis and height equal to the y-coordinate of

P

, which is

1

. So its area is

\frac{1}{2} \cdot \frac{4}{3} \cdot 1 = \frac{2}{3}

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option

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