Question

If the circle $ {{x}^{2}}+{{y}^{2}}+4x+22y+c=0 $ bisects the circumference of the circle $ {{x}^{2}}+{{y}^{2}}-2x+8y-d=0, $ then $ c+d $ is equal to :

• A. 60
• B. 50
• C. 40
• D. 30

Answer 1

Answer: (B)

Given that, circle $x^{2}+y^{2}+4 x+22 y+c=0$
bisects the circumference of the circle
$x^{2}+y^{2}-2 x+8 y-d=0 $
$\therefore $ The common chord of the given circle is
$S_{1}-S_{2}=0 $
$\Rightarrow x^{2}+y^{2}+4 x+22 v+c-x^{2}-y^{2} $
$+2 x-8 y+d=0 $
$\Rightarrow 6 x+14 y+c+d=0$
So, Eq. (i) passes through the centre of the second circle i.e., $(1,-4)$
$\therefore 6-56+c+d=0$
$\Rightarrow c+d=50$
Note : If $S_{1}$ and $S_{2}$ are the equations of two circles,
then equation of common chord is $S_{1}-S_{2}=0$.