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 normal at the point $(3, 4)$ on a circle cuts the circle at the point $(-1, -2)$. Then the equation of the circle is

If the lines $x+2y-5=0$ and $2x-3y+4=0$ lie along diameters of a circle of area $9\pi$,then the equation of the circle is

The radius of the circle $x^2 + y^2 + 4x + 6y + 13 = 0$ is

The circle passing through the point $(-1, 0)$ and touching the $y$-axis at $(0, 2)$ also passes through the point

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module