Tangents are drawn from any point on the circle $x^2 + y^2 = R^2$ to the circle $x^2 + y^2 = r^2$. If the line joining the points of contact of these tangents on the first circle also touches the second circle,then $R$ equals

  • A
    $\sqrt{2} r$
  • B
    $2r$
  • C
    $\frac{2r}{2 - \sqrt{3}}$
  • D
    $\frac{4r}{3 - \sqrt{5}}$

Explore More

Similar Questions

The equation of the pair of tangents at $(0,1)$ to the circle $x^{2}+y^{2}-2x-6y+6=0$ is

Given that $a > 2b > 0$ and that the line $y = mx - b \sqrt{1 + m^2}$ is a common tangent to the circles $x^2 + y^2 = b^2$ and $(x - a)^2 + y^2 = b^2$. Then the positive value of $m$ is

Chords of the curve $4x^2 + y^2 - x + 4y = 0$ which subtend a right angle at the origin pass through a fixed point whose coordinates are:

The distance from the centre of the circle $x^2 + y^2 = 2x$ to the straight line passing through the points of intersection of the two circles $x^2 + y^2 + 5x - 8y + 1 = 0$ and $x^2 + y^2 - 3x + 7y - 25 = 0$ is-

The circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2+2x+4y-11=0$:

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