Statement $(A) :$ If two circles $x^2 + y^2 + 2gx + 2fy = 0$ and $x^2 + y^2 + 2g'x + 2f'y = 0$ touch each other,then $f'g = fg'$.
Reason $(R) :$ Two circles touch each other if the line joining their centers is perpendicular to all possible common tangents.

  • A
    $A$ and $R$ are both independently true and $R$ is the correct explanation for $A$.
  • B
    $A$ and $R$ are both independently true but $R$ is not the correct explanation for $A$.
  • C
    $A$ is true but $R$ is false.
  • D
    $A$ is false but $R$ is true.

Explore More

Similar Questions

If the circles of same radius $a$ and centers at $(2, 3)$ and $(5, 6)$ cut orthogonally,then $a =$

The equation of the direct common tangent of the circles $x^2+y^2-6x-4y-23=0$ and $x^2+y^2+2x+2y+1=0$ is

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

$A$ circle passes through the origin and has its centre on $y = x$. If it cuts ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally,then the equation of the circle is

Difficult
View Solution

The point/points of intersection of the common tangents of the two circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2y-15=0$ 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