A) 30
B) 50
C) 40
D) 56
E) 52
Correct Answer: B
Solution :
Given that,\[{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}+4x+22y+c=0\] bisects the circumference of the circle \[{{S}_{2}}\equiv {{x}^{2}}+{{y}^{2}}-2x+8y-d=0\] The common chord of the given circle is \[{{S}_{1}}-{{S}_{2}}=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+4x+22y+c-{{x}^{2}}-{{y}^{2}}\] \[+2x-8y+d=0\] \[\Rightarrow \] \[6x+14y+c+d=0\] ...(i) So, Eq. (i) passes through the centre of the second circle, ie,\[(1,-4)\]. \[\therefore \] \[6-56+c-4-d=0\] \[\Rightarrow \] \[c+d=50\]
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