Area of triangle formed by tangent, normal drawn at (1,√ 3 ) to circle {x^2} + {y^2} = 4 and posit

English

Gujarati

Hindi

10-1.Circle and System of Circles

medium

The area of triangle formed by the tangent, normal drawn at $(1,\sqrt 3 )$ to the circle ${x^2} + {y^2} = 4$ and positive $x$-axis, is

A

$2\sqrt 3 $

B

$\sqrt 3 $

C

$4\sqrt 3 $

D

None of these

(IIT-1989)

Solution

(a) $T \equiv x + \sqrt 3 y – 4 = 0$

Hence the required area $ = \frac{1}{2} \times 4 \times \sqrt 3 $

$= 2\sqrt 3 $.

Standard 11

Mathematics

Share

Similar Questions

If ${c^2} > {a^2}(1 + {m^2}),$ then the line $y = mx + c$ will intersect the circle ${x^2} + {y^2} = {a^2}$

normal

$y – x + 3 = 0$ is the equation of normal at $\left( {3 + \frac{3}{{\sqrt 2 }},\frac{3}{{\sqrt 2 }}} \right)$ to which of the following circles

medium

The gradient of the normal at the point $(-2, -3)$ on the circle ${x^2} + {y^2} + 2x + 4y + 3 = 0$ is

easy

The square of the length of the tangent from $(3, -4)$ on the circle ${x^2} + {y^2} – 4x – 6y + 3 = 0$ is

easy

The equation of three circles are ${x^2} + {y^2} – 12x – 16y + 64 = 0,$ $3{x^2} + 3{y^2} – 36x + 81 = 0$ and ${x^2} + {y^2} – 16x + 81 = 0.$ The co-ordinates of the point from which the length of tangent drawn to each of the three circle is equal is

hard