The parametric equations of the curve $x^2+y^2-ax-by=0$ are

  • A
    $x=\frac{a}{2}+\sqrt{\frac{a^2+b^2}{4}} \cos \theta, y=-\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin \theta$
  • B
    $x=-\frac{a}{2}+\sqrt{\frac{a^2+b^2}{4}} \cos \theta, y=\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin \theta$
  • C
    $x=-\frac{a}{2}+\sqrt{\frac{a^2+b^2}{4}} \cos \theta, y=-\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin \theta$
  • D
    $x=\frac{a}{2}+\sqrt{\frac{a^2+b^2}{4}} \cos \theta, y=\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin \theta$

Explore More

Similar Questions

The equation of a circle whose centre is the origin and whose radius is equal to the distance between the lines $x = 1$ and $x = -1$ is:

$A$ circle is given by $x^2 + y^2 - 6x + 8y - 11 = 0$. Determine the position of the points $P(0, 0)$ and $Q(1, 8)$ with respect to the circle.

$A$ circle is drawn with the $y$-axis as a tangent and its center at the point which is the reflection of $(3, 4)$ in the line $y = x$. The equation of the circle is:

If the equation of the circle lying in the first quadrant,touching both the coordinate axes and the line $\frac{x}{3}+\frac{y}{4}=1$ is $(x-c)^2+(y-c)^2=c^2$,then $c=$

The equation of a circle passing through the origin and making an $x$-intercept of $3$ and a $y$-intercept of $-5$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo