The equation of the tangent to the circle $x^2+y^2-9=0$ making an angle $60^{\circ}$ with the $X$-axis is

  • A
    $\frac{1}{\sqrt{3}} x-y \pm 6=0$
  • B
    $\sqrt{3} x-y \pm 6=0$
  • C
    $\sqrt{3} x+y \pm 6=0$
  • D
    $\frac{1}{\sqrt{3}} x+y \pm 6=0$

Explore More

Similar Questions

Consider the circle $S: x^2 + y^2 = 1$ and point $P(0, -1)$ on it. $A$ ray of light passes through the point $(-3, -1)$ and reflects from the tangent to $S$ at $P$. After reflection,it becomes tangent to the circle $S$. Find the equation of the reflected ray.

If the lines $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$ are both tangents to a circle,then its radius is

Find the equations of the tangents to the circle $x^2 + y^2 - 6x + 4y = 12$ which are parallel to the line $4x + 3y + 5 = 0$.

The angle between the tangents from $(\alpha, \beta)$ to the circle $x^2 + y^2 = a^2$ is

The slope of the tangent to the circle $(x-6)^2 + y^2 = 2$,which passes through the focus of the parabola $y^2 = 16x$,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