The coordinates of any point on the circle x^ | vedclass

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

Question

(C) The given equation of the circle is $x^2 + y^2 = 4$.This is in the form $x^2 + y^2 = r^2$,where $r^2 = 4$,so the radius $r = 2$.The parametric equ

Vedclass · Accepted Answer

$(2 \cos \alpha, 2 \sin \alpha)$

Vedclass · Answer

(C) The given equation of the circle is $x^2 + y^2 = 4$.This is in the form $x^2 + y^2 = r^2$,where $r^2 = 4$,so the radius $r = 2$.The parametric equations for a circle $x^2 + y^2 = r^2$ are given by $x = r \cos \alpha$ and $y = r \sin \alpha$.Substituting $r = 2$,we get $x = 2 \cos \alpha$ and $y = 2 \sin \alpha$.Thus,the coordinates of any point on the circle are $(2 \cos \alpha, 2 \sin \alpha)$.

The coordinates of any point on the circle $x^2 + y^2 = 4$ are:

Similar Questions

The equation of the circle which touches the $x$-axis and whose centre is $(1, 2)$ is

The equation of a circle which passes through the points $(2, 3)$ and $(4, 5)$ and whose center lies on the straight line $y - 4x + 3 = 0$ is

The equation of the circle whose centre lies on the line $x-4y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is

The equation of the circle that touches the $Y-$axis at a distance of $4$ units from the origin and cuts off an intercept of $6$ units on the $X-$axis is

$A$ circle touches the axes at the points $(3, 0)$ and $(0, -3)$. The centre of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module