If theta is a parameter,then the parametric e | vedclass

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

Question

(B) The given equation of the circle is $x^{2}+y^{2}-6x+4y-3=0$. Completing the square,we get: $(x^{2}-6x+9) + (y^{2}+4y+4) = 3+9+4$ $(x-3)^{2} + (y+2

Vedclass · Accepted Answer

$x=3+4 \cos 	heta$ and $y=-2+4 \sin 	heta$

Vedclass · Answer

(B) The given equation of the circle is $x^{2}+y^{2}-6x+4y-3=0$. Completing the square,we get: $(x^{2}-6x+9) + (y^{2}+4y+4) = 3+9+4$ $(x-3)^{2} + (y+2)^{2} = 16$ $(x-3)^{2} + (y+2)^{2} = 4^{2}$ Comparing this with the standard form $(x-h)^{2} + (y-k)^{2} = r^{2}$,we identify the center $(h, k) = (3, -2)$ and radius $r = 4$. The parametric equations of a circle are given by $x = h + r \cos 	heta$ and $y = k + r \sin 	heta$. Substituting the values,we get $x = 3 + 4 \cos 	heta$ and $y = -2 + 4 \sin 	heta$.

If $\theta$ is a parameter,then the parametric equations of the circle $x^{2}+y^{2}-6x+4y-3=0$ are given by

Similar Questions

The equation of the circle which touches the lines $x = 0$,$y = 0$ and $3x + 4y = 4$ is

$A$ circle has its equation in the form $x^2 + y^2 + 2x + 4y + 1 = 0$. Choose the correct coordinates of its centre and the right value of its radius from the following:

If $(\alpha, \beta)$ is the centre of a circle passing through the origin,then its equation is

The centre and radius of the circle $x^2+y^2+8x+10y-8=0$ respectively are

The equation of the circle of radius $3$ that lies in the fourth quadrant and touches the lines $x=0$ and $y=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module