The area (in sq units) of the triangle formed | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the circle is $x^2+y^2=4$. The point $P(1, \sqrt{3})$ lies on the circle.The equation of the tangent at $P(x_1, y_1)$ is $xx_1 + y

Vedclass · Accepted Answer

$2 \sqrt{3}$

Vedclass · Answer

(C) The equation of the circle is $x^2+y^2=4$. The point $P(1, \sqrt{3})$ lies on the circle.The equation of the tangent at $P(x_1, y_1)$ is $xx_1 + yy_1 = r^2$,which is $x(1) + y(\sqrt{3}) = 4$,or $x + \sqrt{3}y = 4$.The tangent meets the $X$-axis at $A(4, 0)$.The normal at $P(1, \sqrt{3})$ passes through the origin $O(0, 0)$ and has the equation $y = \sqrt{3}x$.The triangle is formed by the points $O(0, 0)$,$P(1, \sqrt{3})$,and $A(4, 0)$.The area of $	riangle OPA = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes OA 	imes y_P = \frac{1}{2} 	imes 4 	imes \sqrt{3} = 2 \sqrt{3}$ sq units.

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

Similar Questions

The length of the tangent drawn to the circle $x^2+y^2-2x+4y-11=0$ from the point $(1,3)$ is:

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

The angle between the tangents from $(\alpha, \beta)$ to the circle $x^2 + y^2 = a^2$ is

Statement $(A) :$ For all values of $\theta$,the line $(x - 3) \cos \theta + (y - 3) \sin \theta = 1$ is tangent to the circle $(x - 3)^2 + (y - 3)^2 = 1$.
Reason $(R) :$ For all values of $\theta$,the line $x \cos \theta + y \sin \theta = a$ is tangent to the circle $x^2 + y^2 = a^2$.

The equation of one of the common tangents of the circle $x^2+y^2-6y+4=0$ and the parabola $y^2=x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module