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 point on the curve ${y^2} = 2(x - 3)$ at which the normal is parallel to the line $y - 2x + 1 = 0$ is

The equation of the normal to the circle $x^2 + y^2 - 2x = 0$ parallel to the line $x + 2y = 3$ is

The equation of the tangent to the circle $x^2+y^2=64$ at the point $P\left(\frac{2\pi}{3}\right)$ is

If $2x - 4y = 9$ and $6x - 12y + 7 = 0$ are the tangents of the same circle,then its radius will be

The line $x = y$ touches a circle at the point $(1, 1)$. If the circle also passes through the point $(1, -3)$,then its radius is

Vedclass Test Series

Exam Paper Generator

Online Exam Module