Question:

A circle \(S\equiv x^2+y^2+4x+2fy+c=0\) passes through the centre of the circle \(x^2+y^2-4x+6y-2=0\). If the line \(3x-3y=c\) passes through the centre of the circle \(S=0\), then the length of tangent from \((1,1)\) to the circle \(S=0\) is

Show Hint

Length of tangent from point \(P\) to circle is \(\sqrt{S_1}\).
Updated On: Jun 17, 2026
  • \(8\)
  • \(\sqrt{2}\)
  • \(5\)
  • \(\sqrt{10}\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Step 1: Find center of second circle.
\[ x^2+y^2-4x+6y-2=0 \] Center: \[ (2,-3) \] Since this lies on \(S\): \[ 4+9+8-6f+c=0 \] \[ 21-6f+c=0 \] \[ c=6f-21 \]

Step 2:
Use center condition.
Center of \(S\): \[ (-2,-f) \] Given: \[ 3x-3y=c \] Substitute: \[ 3(-2)-3(-f)=c \] \[ -6+3f=c \] Equating: \[ 3f-6=6f-21 \] \[ 15=3f \] \[ f=5 \] Thus: \[ c=9 \]

Step 3:
Find tangent length.
Circle: \[ x^2+y^2+4x+10y+9=0 \] At \((1,1)\): \[ 1+1+4+10+9=25 \] Length: \[ =\sqrt{25}=5 \]
Was this answer helpful?
0
0

Top TS EAMCET Coordinate Geometry Questions

View More Questions