A) 60
B) 50
C) 40
D) 30
Correct Answer: B
Solution :
Given that, circle \[{{x}^{2}}+{{y}^{2}}+4x+22y+c=0\] bisects the circumference of the circle \[{{x}^{2}}+{{y}^{2}}-2x+8y-d=0\] \[\therefore \] The common chord of the given circle is \[{{S}_{1}}-{{S}_{2}}=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+4x+22y+c-{{x}^{2}}-{{y}^{2}}\] \[+\,2x-8y+d=0\] \[6x+14y+c+d=0\] ?(i) 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 \[{{S}_{1}}-{{S}_{2}}=0.\]
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