The straight line $x - y - 3 = 0$ touches the circle $x^2 + y^2 - 4x + 6y + 11 = 0$ at the point whose coordinates are

  • A
    $(1, -2)$
  • B
    $(1, 2)$
  • C
    $(-1, 2)$
  • D
    $(-1, -2)$

Explore More

Similar Questions

The point of contact of the tangent to the circle $x^2 + y^2 = 5$ at the point $(1, -2)$ which also touches the circle $x^2 + y^2 - 8x + 6y + 20 = 0$ is:

Difficult
View Solution

The equation of the normal to the circle $2x^2 + 2y^2 - 2x - 5y + 3 = 0$ at the point $(1, 1)$ is:

If the tangents drawn at the points $O(0,0)$ and $P(1+\sqrt{5}, 2)$ on the circle $x^{2}+y^{2}-2x-4y=0$ intersect at the point $Q$,then the area of the triangle $OPQ$ is equal to

If the tangents at the points $P$ and $Q$ on the circle $x^2 + y^2 - 2x + y = 5$ meet at the point $R \left(\frac{9}{4}, 2\right)$,then the area of the triangle $PQR$ is

The angle between the pair of tangents from the point $(1, 1/2)$ to the circle $x^2 + y^2 + 4x + 2y - 4 = 0$ 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