With one focus of the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$ as the centre,a circle is drawn which is tangent to the hyperbola with no part of the circle being outside the hyperbola. The radius of the circle is

  • A
    $less \ than \ 2$
  • B
    $2$
  • C
    $\frac{11}{3}$
  • D
    $none$

Explore More

Similar Questions

If the ellipse $4x^2 + 9y^2 = 36$ is confocal with a hyperbola whose length of the transverse axis is $2$,then the points of intersection of the ellipse and hyperbola lie on the circle:

The equations of common tangents to the parabola $y^2=16x$ and the circle $x^2+y^2=8$ are

Let $PQ$ be a focal chord of the parabola $y^{2}=4x$ such that it subtends an angle of $\frac{\pi}{2}$ at the point $(3, 0)$. Let the line segment $PQ$ be also a focal chord of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$. If $e$ is the eccentricity of the ellipse $E$,then the value of $\frac{1}{e^{2}}$ is equal to

If the curves $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect at right angles,then $a^2 =$

The point$(s)$ on the parabola $y^2 = 4x$ which are closest to the circle $x^2 + y^2 - 24y + 128 = 0$ is/are:

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