A) 41
B) 14
C) 4
D) 0
Correct Answer: C
Solution :
Given circles are \[2{{x}^{2}}+2{{y}^{2}}-3x+6y+k=0\] or \[{{x}^{2}}+{{y}^{2}}-\frac{3}{2}x+3y+\frac{k}{2}=0\] .....(i) and \[{{x}^{2}}+{{y}^{2}}-4x+10y+16=0\] .....(ii) Circle (i) and (ii) cut orthogonally, then \[2{{g}_{1}}\,{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}\] \[2\text{ }\left( -\frac{3}{4} \right)\,(-2)+2\text{ }\left( \frac{3}{2} \right)\,.\,5=\frac{k}{2}+16\] \[3+15=\frac{k}{2}+16\]Þ \[18=\frac{k}{2}+16\]Þ \[k=4.\]
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