A) 41
B) 14
C) 4
D) 1
Correct Answer: C
Solution :
We have given two circles are \[2{{x}^{2}}+2{{y}^{2}}-3x+6y+k=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-\frac{3}{2}x+3y+\frac{k}{2}=0\] ?(i) and \[{{x}^{2}}+{{y}^{2}}-4x+10y+16=0\] ...(ii) Since, general equation of circle is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] ?(iii) Therefore, comparing Eqs. (i) and (ii) with Eq. (iii), we get \[{{g}_{1}}=-\frac{3}{4},\,{{f}_{1}}=\frac{3}{2},\,{{c}_{1}}=\frac{k}{2}\] and \[{{g}_{2}}=-2,\,{{f}_{2}}=5,\,{{c}_{2}}=16\] Since, both the circles cut orthogonally. \[\therefore \] \[2({{g}_{1}}{{g}_{2}}+{{f}_{1}}{{f}_{2}})={{c}_{1}}+{{c}_{2}}\] \[\Rightarrow \] \[2\left( \frac{3}{2}+\frac{15}{2} \right)=\frac{k}{2}+16\] \[\Rightarrow \] \[18=\frac{k}{2}+16\] \[\Rightarrow \] \[\frac{k}{2}=2\] \[\Rightarrow \] \[k=4\]
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