Question

The length of the common chord of the circles \[ x^2+y^2-8x-6y-11=0 \] and \[ x^2+y^2-6x+8y+9=0 \] is

• A. \(\sqrt{42}\)
• B. \(7\)
• C. \(\dfrac{13}{2}\)
• D. \(\sqrt{46}\)

Hint:

The common chord of two circles is their radical axis. Its length is \[ \boxed{2\sqrt{r^2-d^2}}, \] where \(d\) is the perpendicular distance from the centre to the radical axis.

Answer 1

The Correct Option is D

Step 1: Find the radical axis. Subtracting the two circle equations, \[ -2x-14y-20=0, \] or \[ \boxed{x+7y+10=0.} \] This is the common chord.

Step 2: Find the radius of the first circle. The first circle has centre \[ (4,3) \] and radius \[ r=\sqrt{4^2+3^2+11} =\sqrt{36} =6. \]

Step 3: Find the perpendicular distance from the centre to the common chord. The distance from \[ (4,3) \] to \[ x+7y+10=0 \] is \[ d = \frac{|4+21+10|}{\sqrt{1+49}} = \frac{35}{5\sqrt2} = \frac{7}{\sqrt2}. \]

Step 4: Calculate the length of the chord. The chord length is \[ 2\sqrt{r^2-d^2} = 2\sqrt{36-\frac{49}{2}} = 2\sqrt{\frac{23}{2}} = \sqrt{46}. \] Hence, \[ \boxed{\sqrt{46}}. \] Therefore, the correct option is \(\boxed{(D)}\).