A) \[\frac{10}{3}\]
B) \[-\frac{8}{3}\]
C) \[-\frac{10}{3}\]
D) \[\frac{8}{3}\]
Correct Answer: B
Solution :
We have, \[{{x}^{2}}+{{y}^{2}}+kx+4y+2=0\] ... (i) and \[2\left( {{x}^{2}}+{{y}^{2}} \right)-4x-3y+fc=0\] ... (ii) from equation (i) \[\Rightarrow \] \[{{g}_{1}}=k/2,\,{{f}_{1}}=2\]and \[{{c}_{1}}=2\]and from equation (ii) \[\Rightarrow \] \[{{g}_{2}}=-2,\,{{f}_{2}}=-\frac{3}{2},\,{{c}_{2}}=k\] Condition for two circles cut orthogonally is \[i.e.,\] \[2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}\] \[\Rightarrow \] \[2\times \frac{k}{2}\times (-2)+2\times 2\times \left( -\frac{3}{2} \right)=2+k\] \[\Rightarrow \] \[-2\,k-6=2+k\] \[\Rightarrow \] \[-3k=8\] \[\Rightarrow \] \[k=-8/3\]You need to login to perform this action.
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