The centre of the circle given by the paramet | vedclass

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

Question

(B) The given parametric equations are $x = 2 + 3\cos 	heta$ and $y = 3\sin 	heta - 1$.Rearranging these equations to isolate the trigonometric term

Vedclass · Accepted Answer

$(2, -1)$

Vedclass · Answer

(B) The given parametric equations are $x = 2 + 3\cos 	heta$ and $y = 3\sin 	heta - 1$.Rearranging these equations to isolate the trigonometric terms:$x - 2 = 3\cos 	heta \implies \cos 	heta = \frac{x - 2}{3}$$y + 1 = 3\sin 	heta \implies \sin 	heta = \frac{y + 1}{3}$Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we substitute the expressions:$(\frac{y + 1}{3})^2 + (\frac{x - 2}{3})^2 = 1$$(x - 2)^2 + (y + 1)^2 = 3^2$This is the standard form of a circle equation $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the centre.Comparing the equations,the centre $(h, k)$ is $(2, -1)$.

The centre of the circle given by the parametric equations $x = 2 + 3\cos \theta$ and $y = 3\sin \theta - 1$ is

Similar Questions

If a circle touches the lines $3x - 4y - 10 = 0$ and $3x - 4y + 30 = 0$ and its centre lies on the line $x + 2y = 0$,then the equation of the circle is

In $\triangle ABC$,$\angle A = 90^{\circ}$ and the coordinates of points $B$ and $C$ are $(2, -4)$ and $(1, 5)$. Then the equation of the circumcircle of $\triangle ABC$ is

Which of the following is the equation of a circle?

The power of the point $(-3, 7)$ with respect to a circle,with centre $(3, 7)$ and radius $2$,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

Vedclass Test Series

Exam Paper Generator

Online Exam Module