The parametric equations of the circle 2x^2 + | 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 $2x^2 + 2y^2 = 9$. Dividing by $2$,we get $x^2 + y^2 = \frac{9}{2}$. This is in the form $x^2 + y^2 = r^2$,whe

Vedclass · Accepted Answer

$x = \frac{3}{\sqrt{2}} \cos 	heta, y = \frac{3}{\sqrt{2}} \sin 	heta$

Vedclass · Answer

(C) The given equation of the circle is $2x^2 + 2y^2 = 9$. Dividing by $2$,we get $x^2 + y^2 = \frac{9}{2}$. This is in the form $x^2 + y^2 = r^2$,where $r^2 = \frac{9}{2}$,so $r = \frac{3}{\sqrt{2}}$. The parametric equations for a circle $x^2 + y^2 = r^2$ are given by $x = r \cos 	heta$ and $y = r \sin 	heta$. Substituting $r = \frac{3}{\sqrt{2}}$,we get $x = \frac{3}{\sqrt{2}} \cos 	heta$ and $y = \frac{3}{\sqrt{2}} \sin 	heta$.

The parametric equations of the circle $2x^2 + 2y^2 = 9$ are

Similar Questions

The radius of the circle $x^2 + y^2 + 2x \cos \theta + 2y \sin \theta - 8 = 0$ is

The lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area $154$ square units. The equation of the circle is

If the incenter of an equilateral triangle is $(1, 1)$ and the equation of one of its sides is $3x + 4y + 3 = 0$,then the equation of the circumcircle of this triangle is

If the equation $x^{2}+y^{2}-10x+21=0$ has real roots $x=\alpha$ and $y=\beta,$ then

Find the equation of the circle passing through the points $(4, 1)$ and $(6, 5)$ and whose centre is on the line $4x + y = 16$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module