Question:
A circle \(S\) passing through origin cuts another circle
\[
x^2+y^2-6x+8y+16=0
\]
orthogonally and makes a chord of maximum length on line
\[
x-y-2=0
\]
then one diameter of circle \(S\) is
A circle \(S\) passing through origin cuts another circle
\[
x^2+y^2-6x+8y+16=0
\]
orthogonally and makes a chord of maximum length on line
\[
x-y-2=0
\]
then one diameter of circle \(S\) is
Show Hint
Maximum chord of a circle along a line occurs when that line passes through the center.
Updated On: Jun 15, 2026
- \(x+y=2\)
- \(2x+3y=4\)
- \(4x-5y+10=0\)
- \(5x+6y+12=0\)
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The Correct Option is A
Solution and Explanation
Concept:
Two circles cutting orthogonally satisfy:
\[
2g_1g_2+2f_1f_2=c_1+c_2
\]
Maximum chord occurs when line passes through center.
Step 1: Equation of variable circle.
Passing through origin. \[ x^2+y^2+2gx+2fy=0 \]
Step 2: Orthogonal condition.
Given second circle: \[ g=-3,\qquad f=4,\qquad c=16 \] Apply condition \[ 2(g)(-3)+2(f)(4)=16 \] \[ -6g+8f=16 \] \[ -3g+4f=8 \]
Step 3: Maximum chord condition.
Chord maximum when line passes through center. Center of circle \[ (-g,-f) \] Must lie on \[ x-y-2=0 \] So \[ -g+f-2=0 \] \[ f-g=2 \] Solve equations. \[ g=-1,\qquad f=1 \] Diameter line through center and origin: \[ x+y=2 \] Thus \[ \boxed{x+y=2} \]
Step 1: Equation of variable circle.
Passing through origin. \[ x^2+y^2+2gx+2fy=0 \]
Step 2: Orthogonal condition.
Given second circle: \[ g=-3,\qquad f=4,\qquad c=16 \] Apply condition \[ 2(g)(-3)+2(f)(4)=16 \] \[ -6g+8f=16 \] \[ -3g+4f=8 \]
Step 3: Maximum chord condition.
Chord maximum when line passes through center. Center of circle \[ (-g,-f) \] Must lie on \[ x-y-2=0 \] So \[ -g+f-2=0 \] \[ f-g=2 \] Solve equations. \[ g=-1,\qquad f=1 \] Diameter line through center and origin: \[ x+y=2 \] Thus \[ \boxed{x+y=2} \]
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