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 of radius $3$ passes through the point $(7,3)$ and has its centre on the line $x-y-1=0$,then its equation among the following is

Find the equation of the circle with center at $(0,0)$ and radius $r$.

Let the equation $x^{2}+y^{2}+px+(1-p)y+5=0$ represent circles of varying radius $r \in (0, 5]$. Then the number of elements in the set $S = \{q : q = p^{2} \text{ and } q \text{ is an integer}\}$ is ..... .

If the endpoints of a diameter of a circle are $(1, 2)$ and $(3, 4)$,then its equation 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

Vedclass Test Series

Exam Paper Generator

Online Exam Module