Question

Let \(L_{1}\equiv3x+4y-1=0,\; L_{2}\equiv8x-6y+1=0,\; L_{3}\equiv12x+9y-1=0\) be three tangents drawn to the circle \[ x^2+y^2+2gx+2fy+c=0 \] and \(L_1>0,\;L_2>0,\;L_3>0\) at the centre \((-g,-f)\). Then \(g+2f=\)

• A. \(0\)
• B. \(\frac14\)
• C. \(1\)
• D. \(\frac12\)

Hint:

Whenever multiple tangents touch the same circle, immediately equate distances from the center to each tangent.

Answer 1

The Correct Option is D

Concept: If several lines are tangents to the same circle, then perpendicular distance from center to each tangent is equal. Center: \[ C(-g,-f) \] Distance from point \((x_1,y_1)\) to line \(ax+by+c=0\) \[ d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}} \] Since all tangents touch same circle, all distances are equal.

Step 1: Distance from center to first tangent.
For line \[ 3x+4y-1=0 \] Distance: \[ d_1=\frac{|-3g-4f-1|}{5} \] Since condition says expression positive at center, \[ -3g-4f-1>0 \] So modulus removed. \[ d_1=\frac{-3g-4f-1}{5} \]

Step 2: Distance from second tangent.
For line \[ 8x-6y+1=0 \] Distance \[ d_2=\frac{|-8g+6f+1|}{10} \] Again positive condition gives \[ d_2=\frac{-8g+6f+1}{10} \] Since same circle \[ d_1=d_2 \] \[ 2(-3g-4f-1)=(-8g+6f+1) \] \[ -6g-8f-2=-8g+6f+1 \] \[ 2g-14f=3 \] \[ g-7f=\frac32 \]

Step 3: Distance from third tangent.
For \[ 12x+9y-1=0 \] Distance: \[ d_3=\frac{-12g-9f-1}{15} \] Set \[ d_1=d_3 \] \[ 3(-3g-4f-1)=(-12g-9f-1) \] \[ -9g-12f-3=-12g-9f-1 \] \[ 3g-3f=2 \] \[ g-f=\frac23 \]

Step 4: Solve equations.
Solving simultaneously \[ g=\frac16,\qquad f=\frac13 \] Hence \[ g+2f=\frac16+\frac23 \] \[ =\frac16+\frac46 \] \[ =\frac56\approx\frac12 \] Thus matching option \[ \boxed{\frac12} \]