The coordinates of any point on the circle $x^2 + y^2 = 4$ are:
- A$(\cos \alpha, \sin \alpha)$
- B$(4 \cos \alpha, 4 \sin \alpha)$
- C$(2 \cos \alpha, 2 \sin \alpha)$
- D$(\sin \alpha, \cos \alpha)$
Explore More
Similar Questions
The circle $x^2+y^2-8x-12y+\alpha=0$ lies in the first quadrant without touching the coordinate axes. If $(6, 6)$ is an interior point to the circle,then
$A$ circle passes through the centre of another circle $x^2+y^2-3x-4y-1=0$ and its centre is $(5,2)$. Then the equation of this circle is.........
If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area $154 \text{ sq units}$,then the equation of the circle is (Take $\pi = \frac{22}{7}$)
The equation of the circle passing through $(0,0)$ and which makes intercepts $a$ and $b$ on the coordinate axes is
If a circle passing through $(1, -2)$ has $x - y = 2$ and $2x + 3y = 14$ as its diameters,then the radius of the circle is
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo