The equation of the tangent to the circle,giv | vedclass

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

Question

(B) The equation of the tangent to the circle $x^2+y^2=a^2$ at the point $P(	heta)$ is given by $x \cos 	heta + y \sin 	heta = a$. Here,the radius

Vedclass · Accepted Answer

$x+\sqrt{3} y=10$

Vedclass · Answer

(B) The equation of the tangent to the circle $x^2+y^2=a^2$ at the point $P(	heta)$ is given by $x \cos 	heta + y \sin 	heta = a$. Here,the radius $a = 5$ and the angle $	heta = \frac{\pi}{3}$. Substituting these values into the equation: $x \cos \frac{\pi}{3} + y \sin \frac{\pi}{3} = 5$ $x \left( \frac{1}{2} ight) + y \left( \frac{\sqrt{3}}{2} ight) = 5$ Multiplying the entire equation by $2$,we get: $x + \sqrt{3} y = 10$.

The equation of the tangent to the circle,given by $x=5 \cos \theta, y=5 \sin \theta$ at the point $\theta=\frac{\pi}{3}$ on it,is

Similar Questions

The angle between the tangents drawn from the origin to the circle $(x - 7)^2 + (y + 1)^2 = 25$ is:

The equation of the normal at the point $(4, -1)$ of the circle $x^2 + y^2 - 40x + 10y = 153$ is

The angle between the two tangents drawn from the origin to the circle $x^2+y^2-14x+2y+25=0$ is (in $^{\circ}$)

The equations of the tangents to the circle $x^2+y^2=36$ which are perpendicular to the line $5x+y=2$ are:

The line $3x - 2y = k$ meets the circle ${x^2} + {y^2} = 4{r^2}$ at only one point,if ${k^2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module