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
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]
- A
$\frac{1}{{\sqrt 3 }}$
- B
$\frac{4}{{\sqrt 3 }}$
- C
$\frac{1}{3}$
- D
$\frac{2}{{\sqrt 3 }}$
Similar Questions
If the line $lx + my + n = 0$ be a tangent to the circle ${(x - h)^2} + {(y - k)^2} = {a^2},$ then
If the line $lx + my + n = 0$ be a tangent to the circle ${(x - h)^2} + {(y - k)^2} = {a^2},$ then
The equation of the circle having the lines $y^2 - 2y + 4x - 2xy = 0$ as its normals $\&$ passing through the point $(2 , 1)$ is :
The equation of the circle having the lines $y^2 - 2y + 4x - 2xy = 0$ as its normals $\&$ passing through the point $(2 , 1)$ is :
The equation of the chord of the circle ${x^2} + {y^2} = {a^2}$ having $({x_1},{y_1})$ as its mid-point is
The equation of the chord of the circle ${x^2} + {y^2} = {a^2}$ having $({x_1},{y_1})$ as its mid-point is
- [IIT 1983]
The equation of normal to the circle $2{x^2} + 2{y^2} - 2x - 5y + 3 = 0$ at $(1, 1)$ is
The equation of normal to the circle $2{x^2} + 2{y^2} - 2x - 5y + 3 = 0$ at $(1, 1)$ is
A circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is
A circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is