The locus of the point of intersection of the tangents to the circle $x^2+y^2=16$,such that the angle between them is $60^{\circ}$,is

  • A
    $x^2+y^2=4$
  • B
    $x^2+y^2=64$
  • C
    $x^2+y^2=32$
  • D
    $x^2+y^2=48$

Explore More

Similar Questions

If $P(x_1, y_1)$ is a point such that the lengths of the tangents from it to the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ are in the ratio $2:3$,then the locus of $P$ is

The locus of a point,such that the difference of the squares of the lengths of the tangents drawn from it to two given circles is constant,is:

If $A(5, -4)$ and $B(7, 6)$ are points in a plane, then the set of all points $P(x, y)$ in the plane such that $AP:PB = 2:3$ is

If the locus of the midpoints of the chords of the circle $x^2+y^2=25$,which subtend a right angle at the origin,is given by $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$,then $|a|=$

$A$ circle of constant radius $a$ passes through the origin $O$ and cuts the coordinate axes at points $P$ and $Q$. The equation of the locus of the foot of the perpendicular from $O$ to $PQ$ 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