The locus of the centers of the circles which cut the circles $x^2 + y^2 + 4x - 6y + 9 = 0$ and $x^2 + y^2 - 5x + 4y - 2 = 0$ orthogonally is

  • A
    $9x + 10y - 7 = 0$
  • B
    $x - y + 2 = 0$
  • C
    $9x - 10y + 11 = 0$
  • D
    $9x + 10y + 7 = 0$

Explore More

Similar Questions

The equation of the locus of the centroid of the triangle whose vertices are $(a \cos k, a \sin k)$,$(b \sin k, -b \cos k)$ and $(1, 0)$,where $k$ is a parameter,is

If the line $lx + my = 1$ is a tangent to the circle ${x^2} + {y^2} = {a^2}$,then the locus of the point $(l, m)$ is

Difficult
View Solution

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 the centres of the circles,which touch the circle $x^2 + y^2 = 1$ externally,also touch the $y$-axis and lie in the first quadrant is

The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta, 2\sin \theta)$ internally in the ratio $2 : 3$ for all $\theta$ is a

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