A) \[\frac{12}{17}\]
B) \[\frac{12}{17}\]
C) \[-\frac{12}{17}\]
D) \[-\frac{17}{12}\]
Correct Answer: D
Solution :
Given equation of circles \[2{{x}^{2}}+2{{y}^{2}}+4x+5y+1=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+2x+5/2y+1/2=0\] ?..(i) \[3{{x}^{2}}+3{{y}^{2}}+6x-7y+3k=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+2x+7/3+k=0\] ??(ii) Centre and radius of first circle and second circle \[{{C}_{1}}\to (-1,-5/4),\] \[{{R}_{1}}\to \sqrt{1+\frac{25}{16}-1/2}=\sqrt{\frac{16+25-8}{16}}=\sqrt{\frac{33}{16}}\] \[{{C}_{2}}\to (-1,7/6),\] \[{{R}_{2}}\to \sqrt{1+\frac{49}{36}-k}=\sqrt{\frac{85-36k}{36}}\] Now, condition for orthogonality of two circles \[{{({{C}_{1}}{{C}_{2}})}^{2}}={{R}_{1}}^{2}+{{R}_{2}}^{2}\] \[{{(-1+1)}^{2}}+{{(7/6+5/4)}^{2}}=\frac{33}{16}+\frac{85-36k}{36}\] \[0+{{\left( \frac{29}{12} \right)}^{2}}=\frac{33}{16}+\frac{85-36k}{36}\] \[\frac{841}{144}=\frac{33}{16}+\frac{85-36k}{36}\] \[841=297+340-144k\] \[144k=637-841=-204\] \[k=-\frac{17}{12}\]
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