The line $2x - y + 1 = 0$ is tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2y = 4$. The radius of the circle is

  • A
    $3 \sqrt{5}$
  • B
    $5 \sqrt{3}$
  • C
    $2 \sqrt{5}$
  • D
    $5 \sqrt{2}$

Explore More

Similar Questions

The circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ intersect at points $A$ and $B$. Find the equation of the circle with $AB$ as its diameter.

Difficult
View Solution

Let $A$ be a point on the $x$-axis. Common tangents are drawn from $A$ to the curves $x^2+y^2=8$ and $y^2=16x$. If one of these tangents touches the two curves at $Q$ and $R$,then $(QR)^2$ is equal to

If two circles of the same radius $a$ and centers at $(2, 3)$ and $(5, 6)$ are orthogonal,find the value of $a$.

Consider a pair of circles $(|x| - 1)^2 + y^2 = 1$. Ram is moving along the circle centered at $(1, 0)$ in the clockwise direction at a rate of $2 \ m/s$,and Shyam is moving along the circle centered at $(-1, 0)$ in the anticlockwise direction at a rate of $1 \ m/s$. If Ram and Shyam start their journey from the origin $(0, 0)$,then the rate of change of the distance between Ram and Shyam at the instant when Ram crosses the $x$-axis for the first time is:

The set of all values of $a^2$ for which the line $x + y = 0$ bisects two distinct chords drawn from a point $P\left(\frac{1+a}{2}, \frac{1-a}{2}\right)$ on the circle $2x^2 + 2y^2 - (1+a)x - (1-a)y = 0$ is equal to:

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