The coordinates of a point on the circle (x - | 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 $(x - 1)^2 + (y - 1)^2 = 4$.This is in the form $(x - h)^2 + (y - k)^2 = r^2$,where the center $(h, k) = (1, 1

Vedclass · Accepted Answer

$(1 + 2 \cos \alpha, 1 + 2 \sin \alpha)$

Vedclass · Answer

(B) The given equation of the circle is $(x - 1)^2 + (y - 1)^2 = 4$.This is in the form $(x - h)^2 + (y - k)^2 = r^2$,where the center $(h, k) = (1, 1)$ and the radius $r = \sqrt{4} = 2$.The parametric coordinates of a point on a circle with center $(h, k)$ and radius $r$ are given by $(h + r \cos \alpha, k + r \sin \alpha)$.Substituting the values,we get $(1 + 2 \cos \alpha, 1 + 2 \sin \alpha)$.Thus,the correct option is $B$.

The coordinates of a point on the circle $(x - 1)^2 + (y - 1)^2 = 4$ are given by:

Similar Questions

If the equation of the circle having its centre in the second quadrant touches the coordinate axes and also the line $\frac{x}{5}+\frac{y}{12}=1$ is $x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0$,then $\lambda=$

The equation of the smallest circle passing through the points $(2,2)$ and $(3,3)$ is

If the equation of the circle passing through the points $(-1,0), (-1,1), (1,1)$ is $ax^2+ay^2+2gx+2fy-2=0$,then $a=$

If ${g^2} + {f^2} = c$,then the equation ${x^2} + {y^2} + 2gx + 2fy + c = 0$ will represent

The equation of the circle with centre on the $x$-axis,radius $5$ and passing through the point $(2, 3)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module